[Fwd: How Mathematica select one of the roots?]
- To: mathgroup@smc.vnet.net
- Subject: [mg12294] [Fwd: [mg12237] How Mathematica select one of the roots?]
- From: Elvis Dieguez <elvisum@ibm.net>
- Date: Thu, 7 May 1998 18:51:30 -0400
From: Elvis Dieguez <elvisum@ibm.net> To: mathgroup@smc.vnet.net MIME-Version: 1.0 Subject: [mg12294] Re: [mg12237] How Mathematica select one of the roots? References: <199805050729.DAA17059@smc.vnet.net.> Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit If you recall Complex Analysis: (-1)^(1/3) = (Exp[Pi * I + 2*Pi*n*I])^(1/3) where 'n' is any positive integer. The 'principle' value of this expression occurs whenever n = 0. The other two values, in this case, are when n =1 and n = 2. When n =0, you get the answer: Exp[Pi/3 * I] = Cos[Pi/3] + I Sin[Pi/3] = 1/2 + Sqrt[3]/2 * I which is your answer! When you are solving the equation X^1/3 + 1 == 0 you usually want all the solutions which is why Mathematica gives you the values whenever n = 0, 1, 2. Any other questions? See Complex Variables by Churchill (I think that is the author but I am not sure) Elvis Dieguez msm@smol.carrier.kiev.ua wrote: > Dear All! > > I have some question about Mathematica 3.0. > > When Mathematica 3.0 calculate > (-1.)^(1/3) > it generate complex value result > 0.5 + 0.8660254037844385*I. > > But when I try to Solve equation > x^3+1==0 > it generates three different roots. > > So, here is the question: > > "How Mathematica 3.0 select one of the roots in the first example?" --------------94112F1E5D38B2C22A93DC92 Content-Type: message/rfc822 From: Elvis Dieguez <elvisum@ibm.net> To: mathgroup@smc.vnet.net MIME-Version: 1.0 Subject: [mg12294] Re: [mg12237] How Mathematica select one of the roots? References: <199805050729.DAA17059@smc.vnet.net.> Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit If you recall Complex Analysis: (-1)^(1/3) = (Exp[Pi * I + 2*Pi*n*I])^(1/3) where 'n' is any positive integer. The 'principle' value of this expression occurs whenever n = 0. The other two values, in this case, are when n =1 and n = 2. When n =0, you get the answer: Exp[Pi/3 * I] = Cos[Pi/3] + I Sin[Pi/3] = 1/2 + Sqrt[3]/2 * I which is your answer! When you are solving the equation X^1/3 + 1 == 0 you usually want all the solutions which is why Mathematica gives you the values whenever n = 0, 1, 2. Any other questions? See Complex Variables by Churchill (I think that is the author but I am not sure) Elvis Dieguez msm@smol.carrier.kiev.ua wrote: > Dear All! > > I have some question about Mathematica 3.0. > > When Mathematica 3.0 calculate > (-1.)^(1/3) > it generate complex value result > 0.5 + 0.8660254037844385*I. > > But when I try to Solve equation > x^3+1==0 > it generates three different roots. > > So, here is the question: > > "How Mathematica 3.0 select one of the roots in the first example?"