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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


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