       Re: Mathematica strange behaviour finding a cubic root

• To: mathgroup at smc.vnet.net
• Subject: [mg129157] Re: Mathematica strange behaviour finding a cubic root
• From: Murray Eisenberg <murray at math.umass.edu>
• Date: Tue, 18 Dec 2012 02:41:24 -0500 (EST)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• Delivered-to: l-mathgroup@wolfram.com
• Delivered-to: mathgroup-newout@smc.vnet.net
• Delivered-to: mathgroup-newsend@smc.vnet.net
• References: <20121216060645.D7C256924@smc.vnet.net> <20121217075851.C00B8694D@smc.vnet.net>

```Which raises the possibly interesting question: how did Mathematica obtain the result for that FindInstance -- with what would seem unexpectedly large integers in the fractions?

On Dec 17, 2012, at 2:58 AM, Andrzej Kozlowski <akozlowski at gmail.com> wrote:

> 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 at 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:= (-1/2)^(2/3)
>>
>>          1  2/3
>> Out= (-(-))
>>          2
>>
>> In:= N[%]
>>
>> Out= -0.31498 + 0.545562 I
>>
>> In:= ((-1/2)^2)^(1/3)
>>
>>        -(2/3)
>> Out= 2
>>
>> In:= N[%]
>>
>> Out= 0.629961
>>
>>
>> Sergio
>>
>

---
Murray Eisenberg                                    murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower            phone 413 549-1020 (H)
University of Massachusetts                               413 545-2838 (W)
710 North Pleasant Street                         fax   413 545-1801
Amherst, MA 01003-9305

```

• Prev by Date: Re: Do Plot
• Next by Date: Plot Points over an image
• Previous by thread: Re: Mathematica strange behaviour finding a cubic root