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Automatic expression.maxima_methods().rootscontract() ?

asked 2016-03-03 11:39:50 -0600

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I have to compare quite a lot expressions and I like to do it with Sage. Sadly something like

sqrt(x^3)/sqrt(x^2) == sqrt(x)

gives false in default mode sage. Here


does help, but is there a automatic way doing that, at least for one notebook? (Btw, why is it not the default?) Shouldn't at least simplify_full be enough to trigger that?

Maybe worth another question: Are there other pitfalls like this, I should be aware of?

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answered 2016-03-04 11:54:22 -0600

metabeta gravatar image

I think the answer to my actual question should be:


Seems that with these setting one can avoid also other pitfalls when comparing expressions involving sqrts, e.g.


should give True, which it does with the settings, but not without.

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Thanks, that solve my issue.

louisgag gravatar imagelouisgag ( 2016-09-26 10:45:40 -0600 )edit

answered 2016-03-03 11:49:30 -0600

tmonteil gravatar image

updated 2016-03-03 11:51:20 -0600

You should provide some assumptions about x so that the equality becomes true, for example:

sage: assume(x>0)
sage: bool(sqrt(x^3)/sqrt(x^2) == sqrt(x))

Note however that the assumption system is pretty weak.

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Does your comment mean that assumptions are not propagated to a complete expression like here:

sage: %var a,b
sage: assume(a>0)
sage: assume(b>0)
sage: bool(sqrt((a*b)^2)/sqrt(a*b)==sqrt(a*b))

Or am I missing again smth?

metabeta gravatar imagemetabeta ( 2016-03-04 11:09:46 -0600 )edit

Yes, this is the kind of issue with assumptions (and symbolic expression in general). Note that in this case, you can use the canonicalize radical (which makes itself some assumptions, see the doc):

sage: e =  (sqrt((a*b)^2)/sqrt(a*b) == sqrt(a*b))
sage: bool(e.canonicalize_radical())
tmonteil gravatar imagetmonteil ( 2016-03-04 12:00:28 -0600 )edit

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Asked: 2016-03-03 11:39:50 -0600

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Last updated: Mar 04 '16