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 >