On simplifying (a^n)^(1/n)

[Nasser M. Abbasi]

I was looking at some algebra cheat sheet, where it said that
something I never seen before:

(a^n)^(1/n) was "a" if n is odd, and |a| if n is even.

Here is the link

http://tutorial.math.lamar.edu/

It is in the "Algebra Cheat Sheet (Reduced)" little down the page there,
a small pdf file. at the bottom line.

I thought to try it on CAS to see what it says. Maple and
Maxima simplify (a^n)^(1/n) to "a" with no assumptions.

Mathematica will do it only if "n" is assumed to be integers
and also if "a" positive.

Maxima:
----------------
r:a^n$
radcan(r^(1/n));
a

Maple
-----------
restart;
r:=a^n:
simplify(r^(1/n),power,symbolic);
a

Mathematica:
------------ attempt 1---------------
In[27]:= Clear[a, n];
r = a^n;
Simplify[r^(1/n)]
Out[29]= (a^n)^(1/n)

------------ attempt 2---------------
In[30]:= Clear[a, n];
r = a^n;
Assuming[Element[n, Integers] && Element[a, Reals], Simplify[r^(1/n)]]

Out[32]= (a^n)^(1/n)

------------ attempt 3---------------
In[36]:= Clear[a, n];
r = a^n;
Assuming[a > 0, Simplify[r^(1/n)]]

Out[38]= (a^n)^(1/n)

------------ attempt 4---------------
In[33]:= Clear[a, n];
r = a^n;
Assuming[Element[n, Integers] && a > 0, Simplify[r^(1/n)]]

Out[35]= a
-----------------------------

Question is: Math wise, which CAS is correct? Is the cheat sheet
correct in saying
(a^n)^(1/n) was "a" if n is odd, and |a| if n is even.

What does your CAS do for this one?

--Nasser

[Axel Vogt]

On 12.09.2016 06:25, Nasser M. Abbasi wrote:
> I was looking at some algebra cheat sheet, where it said that
> something I never seen before:
>
> (a^n)^(1/n) was "a" if n is odd, and |a| if n is even.

...

> Maple
> -----------
> restart;
> r:=a^n:
> simplify(r^(1/n),power,symbolic);
> a

Be aware that this is a false instruction for Maple, 'symbolic'
tells Maple to ignore (some) rules.

# Maple 2016:

(a^n)^(1/n): subs(a=-1, n=2, %);
1

(a^n)^(1/n): subs(a=-1, n=3, %): simplify(%);

1/2
1/2 + 1/2 I 3

[Axel Vogt]

On 12.09.2016 06:25, Nasser M. Abbasi wrote:

> I was looking at some algebra cheat sheet, where it said that
> something I never seen before:
>
> (a^n)^(1/n) was "a" if n is odd, and |a| if n is even.

...

The *general* formula (which is more usefull, I think) is:

(z^a)^b = z^(a*b) * exp(-2*Pi*I*b*K(a*ln(z)))

where K = "unwinding number" (which cares for brunch cuts
in a consistent (!) way),

K(z) = ceil(1/2*(Im(z)-Pi)/Pi) # Maple
= 1/2*I*(-z+ln(exp(z)))/Pi # "Blätterzahl"
= 1/2*(Im(z)-argument(exp(z)))/Pi # Aslaksen (*)

Note that brunch cuts for log and sqrt need to fit (there
is neat theorem connecting log and sqrt in complex analysis).

(*) Aslaksen: Can your computer do complex analysis (2005)