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Re: Mathematica strange behaviour finding a cubic root
Posted:
Dec 17, 2012 2:54 AM


You can't because they are not equal of course. Fractional powers are defined as
x^a = Exp[a*Log[x]
where Log is the principal branch of the logarithm. It is impossible to define a continuous branch of the logarithm in the entire complex plane so as you go around it there has to be a "jump" somewhere. The usual choice of the so called principal branch makes the jump take place on the negative real axis. The two answers that you get to yoru computation come from different branches of the logarithm. In fact here is one of your answers:
Exp[(1/3)*Log[1/4]]
1/2^(2/3)
and here is the other:
Simplify[Exp[(1/3)*(Log[1/4] + 2*Pi*I)]]
((1/2))^(2/3)
they are certainly not equal. The reason why you think they are equal is because you are assuming that
(x^a)^b = x^(a b)
but this is not always true. In fact Mathematica itself can find examples when this does not hold, e.g:
FindInstance[x^(a*b) != (x^a)^b && Element[{x, b}, Reals] && Element[a, Integers], {x, a, b}]
{{x > (109/5), a > 22, b > (56/5)}}
Andrzej Kozlowski
On 16 Dec 2012, at 07:06, sergio_r@mail.com wrote:
> > How can I make Mathematica provides the same answer for > (1/2)^(2/3) = ((1/2)^2)^(1/3) ? > > What follows is a Mathematica session: > > In[1]:= (1/2)^(2/3) > > 1 2/3 > Out[1]= (()) > 2 > > In[2]:= N[%] > > Out[2]= 0.31498 + 0.545562 I > > In[3]:= ((1/2)^2)^(1/3) > > (2/3) > Out[3]= 2 > > In[4]:= N[%] > > Out[4]= 0.629961 > > > Sergio >



