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
- References:
- Cube root of -1 and 1
- From: Bob F <deepyogurt@gmail.com>
- Cube root of -1 and 1