Re: Cube root of -1 and 1
- To: mathgroup at smc.vnet.net
- Subject: [mg91028] Re: Cube root of -1 and 1
- From: "David Park" <djmpark at comcast.net>
- Date: Mon, 4 Aug 2008 03:22:23 -0400 (EDT)
- References: <g6h5fm$h26$1@smc.vnet.net>
Mathematica just does a straight forward calculation and does not try to
find multiple roots.
(Sqrt[36] - 7)^(1/3) // Trace
{{{Sqrt[36],6},6-7,-1},{{1/3,1/3},1/3,1/3},(-1)^(1/3)}
(Sqrt[36]-5)^(1/3)//Trace
{{{Sqrt[36],6},6-5,1},{{1/3,1/3},1/3,1/3},1^(1/3),1}
(Notice that 1^x gives 1.)
If you want all the roots use Solve:
Solve[w^3 == (Sqrt[36] - 7)]
{{w->-1},{w->(-1)^(1/3)},{w->-(-1)^(2/3)}}
Solve[w^3 == (Sqrt[36] - 5)]
{{w->1},{w->-(-1)^(1/3)},{w->(-1)^(2/3)}}
--
David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/
"Bob F" <deepyogurt at gmail.com> wrote in message
news:g6h5fm$h26$1 at smc.vnet.net...
> 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
>