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Re: Cube root of -1 and 1
*To*: mathgroup at smc.vnet.net
*Subject*: [mg90893] Re: [mg90875] Cube root of -1 and 1
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Mon, 28 Jul 2008 07:53:46 -0400 (EDT)
*References*: <200807270632.CAA17232@smc.vnet.net>
On 27 Jul 2008, at 08:32, Bob F wrote:
> Could someone explain why Mathematica evaluates these so differently?
>
> In[53]:=
>
> (Sqrt[36] - 7)^(1/3)
> (Sqrt[36] - 5)^(1/3)
>
> Out[53]= (-1)^(1/3)
>
> Out[54]= 1
>
> In other words why isn't (-1)^1/3 expressed as -1 ??
>
> Thanks...
>
> -Bob
>
The function z->z^n is a complex multivalued function, and, of course,
there are three values that could serve as (-1)^(1/3). The most
convenient choice is the so called "principal value", which in case is
not -1. To see this clearly, note that Mathematica uses the following
definition of Power:
Power[x,y] = Exp[y Log[x]]
So the choice of the principal value of Power[-1,1/3] amounts to the
choice principal value of Log[-1]. What should be that?
Well, we would like the following relationship between Log and Arg to
hold:
Log[z] = log[Abs[z]]+I Arg[z]
hence
Log[-1]= I Arg[-1]
Since clearly one would not want Arg to be anything else but:
Arg[-1]
Pi
we have to choose Log[-1] as Pi I and hence Power[-1,1/3] as Exp[Pi I/
3]. Any other choice would make many formulas and computations more
complicated.
Andrzej Kozlowski
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