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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


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