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[?]  Simplify powers, (x^n)^(1/n) to x?

Kenneth Miller
Posted 8 years ago

Mathematica fails to simplify the following, even with the Assumptions I give it:

In[59]:= FullSimplify[(xe^n)^(1/n), 
 Assumptions -> {xe \[Element] Reals, n \[Element] integers, n > 0}]

Out[59]= (xe^n)^(1/n)

Does anyone know how to get Mathematica to simplify (x^n)^(1/n) to x?
POSTED BY: Kenneth Miller
4 Replies
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Hans Milton
Hans Milton
Posted 8 years ago

PowerExpand applied to complex numbers:

In[1]:= 
  z = a + I b;
  w = c + I d;

In[3]:= u = (z^w)^(1/w)

Out[3]= ((a + I b)^(c + I d))^(1/(c + I d))

In[4]:= PowerExpand@u

Out[4]= a + I b
POSTED BY: Hans Milton
Hans Milton
Hans Milton
Posted 8 years ago

You are of course right Gianluca. PowerExpand acts here like a quick and easy solution for everyday use. When the variables are real and positive numbers.

Given the right assumptions Simplify will also give the expected result:

In[1]:= Simplify[(x^n)^(1/n), Assumptions -> {x \[Element] Reals, x > 0, n \[Element] Reals}]

Out[1]= x
POSTED BY: Hans Milton

Your simplification of (x^n)^(1/n) to x is correct for x>0, but it is wrong or problematic for x<0, or for complex x. See what happens for x=-2, n=2:

With[{x=-2, n=2}, {(x^n)^(1/n), x}]

Mathematica refuses to make the simplification unless the user explicitly assumes that x is positive. PowerExpand instructs to ignore all those scruples.
POSTED BY: Gianluca Gorni
Hans Milton
Hans Milton
Posted 8 years ago

PowerExpand does it

In[2]:= PowerExpand[(x^n)^(1/n)]

Out[2]= x
POSTED BY: Hans Milton
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