Re: principle root? problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg32543] Re: principle root? problem*From*: "Allan Hayes" <hay at haystack.demon.co.uk>*Date*: Sat, 26 Jan 2002 04:08:04 -0500 (EST)*References*: <a2r4jh$9p4$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Rob, > Now, is this a > "principle root" problem and are the two possibly related? Yes to both questions; though you may have meant x^(1/2)= 16, not x^(1/2)= -16 , which Mathematica says has no solution. To Mathematica, z^a , when z is not 0, is the principle value: Exp[a(Log[Abs[z] + I Arg[z])] where Log[Abs[z]] is the ordinary logarithm of the real number Abs[z] and Arg[z] is the principle argument of z, the one lying in (-Pi, Pi] (including Pi but not -Pi). One would automatically think of changing the first equation to Solve[(x^2 - 5)^2 == 16^3, x] {{x -> (-I)*Sqrt[59]}, {x -> I*Sqrt[59]}, {x -> -Sqrt[69]}, {x -> Sqrt[69]}} The following way of making Mathematica look at all values, not just the principle value, works for rational powers: Solve[{u^2 == 16, u^3 == x^2 - 5}, x] {{x -> (-I)*Sqrt[59]}, {x -> I*Sqrt[59]}, {x -> -Sqrt[69]}, {x -> Sqrt[69]}} Solve[{u == -16, x == u^2}, {u, x}] {{x -> 256, u -> -16}} -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 "RDownes" <rdownes at aol.com> wrote in message news:a2r4jh$9p4$1 at smc.vnet.net... > The other day, I was explaining something to a student regarding solving a > simple algebra problem. > > (x^2-5)^(2/3)=16 > > The solution to which is easily found. All four that is! However my version of > Mathematica only gave the two real. Is there a simple explanation for this? > > Also, Mathematica gives the solution to x^(1/2)= -16 as 256. Now, is this a > "principle root" problem and are the two possibly related? Any insights would > be appreciated for this little enigma. > > Thanks, > > Rob > >