       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-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:= Assuming[x > 0, Simplify[0^x]]

Out= 0

David

> Zero to the power of zero is Indeterminate.
>
> Steven Siew
>
> > In:=
> >
> > (Sqrt - 7)^(1/3)
> > (Sqrt - 5)^(1/3)
> >
> > Out= (-1)^(1/3)
> >
> > Out= 1
> >
> > In other words why isn't (-1)^1/3 expressed as -1 ??
> >
> > Thanks...
> >
> > -Bob

```

• Prev by Date: Re: Warning for use of 3D Graphics manipulation with notebook
• Next by Date: Re: bug? f'[x]'
• Previous by thread: Re: Cube root of -1 and 1
• Next by thread: Method Option