Re: Cube root of -1 and 1
- To: mathgroup at smc.vnet.net
- Subject: [mg90883] Re: [mg90875] Cube root of -1 and 1
- From: Murray Eisenberg <murray at math.umass.edu>
- Date: Mon, 28 Jul 2008 07:51:53 -0400 (EDT)
- Organization: Mathematics & Statistics, Univ. of Mass./Amherst
- References: <200807270632.CAA17232@smc.vnet.net>
- Reply-to: murray at math.umass.edu
Because of the following? Arg[-1] Pi Arg[1] 0 According to the reference page for Power, "For complex numbers x and y, Power gives the principal value of e^(y log (x))." And an example there shows the result (-1)^(1/3) for (-1)^(1/3) and says, "The principal root is always used." If you want an actual expansion for (-1)^(1/3) into real and complex parts, then use ComplexExpand: ComplexExpand[(-1)^(1/3)] // InputForm 1/2 + (I/2)*Sqrt[3] 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 > -- Murray Eisenberg murray at math.umass.edu Mathematics & Statistics Dept. Lederle Graduate Research Tower phone 413 549-1020 (H) University of Massachusetts 413 545-2859 (W) 710 North Pleasant Street fax 413 545-1801 Amherst, MA 01003-9305
- References:
- Cube root of -1 and 1
- From: Bob F <deepyogurt@gmail.com>
- Cube root of -1 and 1