Hello,
In Mathemtiaca
Simplify[Sqrt[(a b c + d e g)^2], a b c + d e g > 0]
will produce
Abs[a b c + d e g]
It is a little wired, as I think the answer should be
a b d+ d e g
Just my guess. Simplify transforms the condition a*b*c + d*e*f>0 into the complicated result of Reduce[a*b*c + d*e*f>0] and then gets bogged down. It does not recognize that it fits the pattern Simplify[Sqrt[x^2], x > 0]. Funny that the following does work:
With[{x = Sin[a*b*c + d*e*f ]},
Simplify[Sqrt[x^2], x > 0]]
Yes. It is fuuny. Is there a way to
Simplify[Sqrt[(a b c + d e g)^2], a b c + d e g > 0]
The easiest would just be removing the absolutes:
yourexpression /. Abs -> Identity
Does this help?
Assuming[a b c + d e g > 0, Simplify[Sqrt[(a b c + d e g)^2]]]
Out[22]= Abs[a b c + d e g]
Never mind; that’s what you started with!
Misread OP.
Deleted
Simplify[Sqrt[(a b c + d e g)^2], a b c>0&&d e g>0]
returns
a b c+d e g
You might be able to loosen that up a little bit by indicating the result is still positive if one of those triples is negative
Your example shows the square of the root. In that case no domain specification is needed. The OP asked for the root of the square,
Try this:
Simplify[Sqrt[(a b c+d e f)^2],(a b c+d e f)\[Element]NonNegativeReals]
It gives:
a b c + d e f
You are of course correct. Apologies.
Your way is very good. Do you feel a lilttle weird as \[Element] NonNegativeReals works but >0 doesn’t?
May you tell why does this work? I don’t know the reason. For example, this doesn’t work
List[{a, b, c, d}] /. List -> Identity
Well, looking at this further I see that this would also work:
Simplify[Sqrt[(a b c + d e f)^2], (a b c + d e f) >= 0]
So zero needs to be included in the domain of this case.
But when the expression is shorter there is no such need:
Simplify[Sqrt[(a b + d e)^2], (a b + d e) > 0]
This is puzzling
Sqrt[(a b c + d e g)^2] // PowerExpand
If in this situation
Simplify[Sqrt[1/(a b c + d e f)^2], (a b c + d e f) > 0]
we can’t use
>=0
here.
Weird. Your way works.