Simplify square root of square

[Paul Royik]

What is the way to consistently simplify square roots of squares?

Examples:

sqrt((x+1)^2) - > x+1

sqrt(cos(4*x)+1) -> sqrt(2)cos(2x)

[Simon King]

Hi Paul,

Simplification must not change the value of the expression. sqrt(x^2) is
certainly not equal to x. Even under the additional assumption that x is
real, you could only simplify it to abs(x)

Best regards,
Simon

[Vincent Delecroix]

But...

sage: eq = sqrt((pi-5)^2)
sage: eq.canonicalize_radical()
pi - 5

And as you can read from the documentation
"""
Choose a canonical branch of the given expression. The square root,
cube root, natural log, etc. functions are multi-valued. The
"canonicalize_radical()" method will choose *one* of these values
based on a heuristic.
"""

As Simon said, be careful if you start using canonicalize_radical
since it does not preserve equality

sage: eq = sqrt((pi-5)^2)
sage: eq.numerical_approx()
1.85840734641021
sage: eq.canonicalize_radical().numerical_approx()
-1.85840734641021

Vincent

2015-02-27 14:36 UTC+01:00, Simon King <simon...@uni-jena.de>:

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[Paul Royik]

OK. Let x is real.

How to rewrite sqrt(cos(4x)+1) into sqrt(2)abs(cos(2x))?

[Vincent Delecroix]

Here is one way... not sure it is the best

sage: eq1 = sqrt(cos(4*x)+1)
sage: eq2 = eq1.simplify_trig()
sage: eq2
sqrt(8*cos(x)^4 - 8*cos(x)^2 + 2)

The next step consists in factoring what is inside the sqrt:

sage: o = eq2.operands()[0]
sage: of = o.factor()
sage: o
8*cos(x)^4 - 8*cos(x)^2 + 2
sage: of
2*(2*cos(x)^2 - 1)^2
sage: eq3 = eq2.subs({o:of})
sage: eq3
sqrt(2)*sqrt((2*cos(x)^2 - 1)^2)

Then for the conversion sqrt(x^2) -> abs(x) you can use

sage: eq3.simplify_real()
sqrt(2)*abs(2*cos(x)^2 - 1)

Vincent

[Paul Royik]

Thank you!