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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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