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
>
>