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Topic: Mathematica strange behaviour finding a cubic root
Replies: 5   Last Post: Dec 18, 2012 2:36 AM

 Messages: [ Previous | Next ]
 Andrzej Kozlowski Posts: 226 Registered: 1/29/05
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

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
>

Date Subject Author
12/17/12 Bob Hanlon
12/17/12 Murray Eisenberg
12/17/12 Andrzej Kozlowski
12/18/12 Murray Eisenberg