Could someone explain why these two expressions are not equivalent?
Sqrt[(a^2 + b^2)^7] === (a^2 + b^2)^(7/2)
Thanks.
=== checks for equality of expressions. Compare
===
FullForm[Sqrt[(a^2 + b^2)^7]]
Power[Power[Plus[Power[a, 2], Power[b, 2]], 7], Rational[1, 2]]
with
FullForm[(a^2 + b^2)^(7/2)]
Power[Plus[Power[a, 2], Power[b, 2]], Rational[7, 2]]
Okay. Agreed. The way Mathematica parses these expressions is different. But algebraically these two expressions are equivalent. Is there anyway to expand them to demonstrate this fact? I've tried
Sqrt[(a^2+b^2)^7//Expand] and Expand[(a^2+b^2)^(7/2)] with no luck. Thoughts? (And thanks).
(By simply checking (3^2+4^2)^(7/2) and Sqrt[(3^2+4^2)]^7 I, of course, get the same answer)
PowerExpand[Sqrt[(a^2 + b^2)^7]] === PowerExpand[(a^2 + b^2)^(7/2)]
Great. Thanks very much.
Your expressions are not equivalent over the complexes:
Sqrt[(a^2 + b^2)^7] == (a^2 + b^2)^(7/2) /. {a -> I, b -> I} {Sqrt[(a^2 + b^2)^7], (a^2 + b^2)^(7/2)} /. {a -> I, b -> I}
I used assumptions that both a and b are non-negative real numbers. I still needed to use PowerExpand to make it work. Thanks for the feedback.
Using Simplify with assumptions :
Simplify
Simplify[{Sqrt[(a^2+b^2)^7],(a^2+b^2)^(7/2)},Element[{a,b},NonNegativeReals]]
Ohh.. I like that. A better demonstration. Thanks Hans Milton.
