Re: Cube root of -1 and 1

*To*: mathgroup at smc.vnet.net*Subject*: [mg90892] Re: Cube root of -1 and 1*From*: Peter Breitfeld <phbrf at t-online.de>*Date*: Mon, 28 Jul 2008 07:53:35 -0400 (EDT)*Organization*: SFZ Bad Saulgau*References*: <g6h5fm$h26$1@smc.vnet.net>

Bob F schrieb: > 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 > Mathematica treats every number as a complex one. So it returns always the principal value, the value with the smallest Arg. Your first example is (-1)^(1/3) which is multivalued with the possible values {Exp[I 60°], -1=Exp[I 180°], Exp[I 300°]}. The smalles Arg is 60°, so this one is given. In the second example (1)^(1/3) the set is {1=Exp[I 0°], Exp[I 120°], Exp[I 240°]}, so 0° is the smallest and you get -1. Gruss Peter -- ==-==-==-==-==-==-==-==-==-==-==-==-==-==-==-==-==-==-==-==-==-== Peter Breitfeld, Bad Saulgau, Germany -- http://www.pBreitfeld.de