Re: Cube root of -1 and 1

*To*: mathgroup at smc.vnet.net*Subject*: [mg90899] Re: Cube root of -1 and 1*From*: "David W.Cantrell" <DWCantrell at sigmaxi.net>*Date*: Tue, 29 Jul 2008 01:36:53 -0400 (EDT)*References*: <g6h5fm$h26$1@smc.vnet.net> <g6kc35$k5t$1@smc.vnet.net>

Steven Siew <stevensiew2 at gmail.com> wrote: > On Jul 27, 4:44 pm, Bob F <deepyog... at gmail.com> wrote: > > Could someone explain why Mathematica evaluates these so differently? > > > Try this > > Table[(Sqrt[36]-n)^(1/3),{n,0,7}] > > (1)^(1/3) > > (0)^(1/3) > > The short answer is : > > One to the power of anything is one > > Zero to the power of anything (non-zero) is zero. No. Rather, zero to any _positive_ power is zero. In[5]:= Assuming[x > 0, Simplify[0^x]] Out[5]= 0 David > Zero to the power of zero is Indeterminate. > > Steven Siew > > > 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