# How to simplify Sqrt[1/x] Sqrt[x]?

In my expression, there appear terms of the form A^(B Sqrt[1/C] Sqrt[C]). Mathematica doesn't realize that this is just simply A^B. I tried telling it explicitly by some replacement rule. This works for simple cases, but somehow if the form above is embedded in a larger expression it does not do this replacement rule.

For example,

(9 + 9 E^((4 t Sqrt[Λ])/Sqrt[3]) -
12 y^2 Λ - 12 z^2 Λ +
6 y^2 Sqrt[1/Λ] Λ^(3/2) +
6 z^2 Sqrt[1/Λ] Λ^(3/2) +
x^4 Λ^2 + y^4 Λ^2 +
2 y^2 z^2 Λ^2 + z^4 Λ^2 +
2 x^2 Λ (-6 +
3 Sqrt[1/Λ]
Sqrt[Λ] + (y^2 + z^2) Λ) +
6 E^((2 t Sqrt[Λ])/Sqrt[
3]) (3 - 2 z^2 Λ +
x^2 (-2 +
Sqrt[1/Λ]
Sqrt[Λ]) Λ +
y^2 (-2 +
Sqrt[1/Λ]
Sqrt[Λ]) Λ +
z^2 Sqrt[1/Λ] Λ^(3/2)))/(9 +
9 E^((4 t Sqrt[Λ])/Sqrt[3]) -
12 y^2 Λ - 12 z^2 Λ +
6 y^2 Sqrt[1/Λ] Λ^(3/2) +
6 z^2 Sqrt[1/Λ] Λ^(3/2) +
x^4 Λ^2 + y^4 Λ^2 +
2 y^2 z^2 Λ^2 + z^4 Λ^2 +
2 x^2 Λ (-6 +
3 Sqrt[1/Λ]
Sqrt[Λ] + (y^2 + z^2) Λ) -
6 E^((2 t Sqrt[Λ])/Sqrt[
3]) (-3 + x^2 Sqrt[1/Λ] Λ^(3/2) +
y^2 Sqrt[1/Λ] Λ^(3/2) +
z^2 Sqrt[1/Λ] Λ^(3/2)))/. A_^(B_. Sqrt[1/Λ] Sqrt[Λ]) :> A^B


Could anyone help me?

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is this what you mean? Assuming[x > 0, Simplify[Sqrt[1/x] Sqrt[x]]] !Mathematica graphics and Assuming[x > 0, Simplify[A^(B*Sqrt[1/x] Sqrt[x])]] gives $A^B$ – Nasser Aug 20 '13 at 18:37
PowerExpand[A^(B Sqrt[1/C] Sqrt[C])]..by default M takes in consideration Complex numbers too. On forcing assumption or PowerExpand it works without Complexes. – Rorschach Aug 20 '13 at 18:38

That is only true if C (in your short example) is positive, therefore you must instruct Mathematica to make such assumptions:

FullSimplify[expr, Λ > 0]


1

You could also do this with $Assumptions: $Assumptions = {Λ > 0};

FullSimplify[expr]


1

Blackbird suggests:

PowerExpand[expr] // FullSimplify


1

Where:

expr = (9 + 9 E^((4 t Sqrt[Λ])/Sqrt[3]) - 12 y^2 Λ -
12 z^2 Λ + 6 y^2 Sqrt[1/Λ] Λ^(3/2) +
6 z^2 Sqrt[1/Λ] Λ^(3/2) + x^4 Λ^2 +
y^4 Λ^2 + 2 y^2 z^2 Λ^2 + z^4 Λ^2 +
2 x^2 Λ (-6 +
3 Sqrt[1/Λ] Sqrt[Λ] + (y^2 +
z^2) Λ) +
6 E^((2 t Sqrt[Λ])/Sqrt[3]) (3 - 2 z^2 Λ +
x^2 (-2 + Sqrt[1/Λ] Sqrt[Λ]) Λ +
y^2 (-2 + Sqrt[1/Λ] Sqrt[Λ]) Λ +
z^2 Sqrt[1/Λ] Λ^(3/2)))/(9 +
9 E^((4 t Sqrt[Λ])/Sqrt[3]) - 12 y^2 Λ -
12 z^2 Λ + 6 y^2 Sqrt[1/Λ] Λ^(3/2) +
6 z^2 Sqrt[1/Λ] Λ^(3/2) + x^4 Λ^2 +
y^4 Λ^2 + 2 y^2 z^2 Λ^2 + z^4 Λ^2 +
2 x^2 Λ (-6 +
3 Sqrt[1/Λ] Sqrt[Λ] + (y^2 +
z^2) Λ) -
6 E^((2 t Sqrt[Λ])/Sqrt[3]) (-3 +
x^2 Sqrt[1/Λ] Λ^(3/2) +
y^2 Sqrt[1/Λ] Λ^(3/2) +
z^2 Sqrt[1/Λ] Λ^(3/2)));

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Thanks! Can I also define $Lambda$ globally to be positive? – user25477 Aug 20 '13 at 18:51
@user25477: question is..shall you ? – Rorschach Aug 20 '13 at 18:52
I want to... but how? – user25477 Aug 20 '13 at 18:56
@user25477 See my update; use \$Assumptions – Mr.Wizard Aug 20 '13 at 19:35