Re: Re: FindRoot
- To: mathgroup at smc.vnet.net
- Subject: [mg98753] Re: [mg98744] Re: [mg98693] FindRoot
- From: DrMajorBob <btreat1 at austin.rr.com>
- Date: Sat, 18 Apr 2009 03:36:42 -0400 (EDT)
- References: <200904170830.EAA23546@smc.vnet.net>
- Reply-to: drmajorbob at bigfoot.com
To see real roots that APPEAR real (when a is real), I'd use something like
eqn = x^3 - 4 a^2 x - 2 a^3 == 0;
r = Simplify[ComplexExpand[x /. Solve[eqn, x]], a > 0]
{(4 a Cos[1/3 ArcTan[Sqrt[37/3]/3]])/Sqrt[3], -(2/3)
a (Sqrt[3] Cos[1/3 ArcTan[Sqrt[37/3]/3]] +
3 Sin[1/3 ArcTan[Sqrt[37/3]/3]]),
a (-((2 Cos[1/3 ArcTan[Sqrt[37/3]/3]])/Sqrt[3]) +
2 Sin[1/3 ArcTan[Sqrt[37/3]/3]])}
Plot[r, {a, -5, 5}]
or equivalently:
eqn2 = eqn /. x -> a y;
r = y /. ComplexExpand@Solve[eqn2, y]
{(4 Cos[1/3 ArcTan[Sqrt[37/3]/3]])/Sqrt[3], -((
2 Cos[1/3 ArcTan[Sqrt[37/3]/3]])/Sqrt[3]) +
2 Sin[1/3 ArcTan[Sqrt[37/3]/3]], -((
2 Cos[1/3 ArcTan[Sqrt[37/3]/3]])/Sqrt[3]) -
2 Sin[1/3 ArcTan[Sqrt[37/3]/3]]}
Plot[a r, {a, -5, 5}]
Bobby
On Fri, 17 Apr 2009 03:30:30 -0500, Bob Hanlon <hanlonr at cox.net> wrote:
> eqn = x^3 - 4 a^2 x - 2 a^3 == 0;
>
> r = Simplify[x /. Solve[eqn, x], a > 0]
>
> {((4*3^(1/3) + (9 + I*Sqrt[111])^
> (2/3))*a)/(3^(2/3)*
> (9 + I*Sqrt[111])^(1/3)),
> (I*(-12 + 4*I*Sqrt[3] +
> I*3^(1/6)*(9 + I*Sqrt[111])^
> (2/3) + (27 + 3*I*Sqrt[111])^
> (2/3))*a)/(2*3^(5/6)*
> (9 + I*Sqrt[111])^(1/3)),
> ((12*I - 4*Sqrt[3] -
> 3^(1/6)*(9 + I*Sqrt[111])^
> (2/3) -
> I*(27 + 3*I*Sqrt[111])^(2/3))*
> a)/(2*3^(5/6)*(9 + I*Sqrt[111])^
> (1/3))}
>
> Plot[r, {a, 0, 5}]
>
>
> Bob Hanlon
>
> ---- Miguel <misvrne at gmail.com> wrote:
>
> =============
> Hi all,
> How can I to find the reals roots of a cubic equation in simbolic
> form: For example,
>
> Find the roots of
> x^3-4a^2x-2a^3==0
>
> where "a" is real and a>0.
>
> Thanks
>
>
>
--
DrMajorBob at bigfoot.com
- References:
- Re: FindRoot
- From: Bob Hanlon <hanlonr@cox.net>
- Re: FindRoot