Re: Bug in Solve?
- To: mathgroup at smc.vnet.net
- Subject: [mg102973] Re: [mg102921] Bug in Solve?
- From: DrMajorBob <btreat1 at austin.rr.com>
- Date: Thu, 3 Sep 2009 05:41:46 -0400 (EDT)
- References: <200909020803.EAA03289@smc.vnet.net>
- Reply-to: drmajorbob at yahoo.com
Three real roots:
f[x_] := x^3 - 15 x + 2
x /. Solve[f[x] == 0, x] // ComplexExpand // Simplify
{Sqrt[5] (Cos[1/3 ArcTan[2 Sqrt[31]]] +
Sqrt[3] Sin[1/3 ArcTan[2 Sqrt[31]]]), -2 Sqrt[5]
Cos[1/3 ArcTan[2 Sqrt[31]]],
Sqrt[5] (Cos[1/3 ArcTan[2 Sqrt[31]]] -
Sqrt[3] Sin[1/3 ArcTan[2 Sqrt[31]]])}
Bobby
On Wed, 02 Sep 2009 03:03:32 -0500, tonysin <a2mgoog at yahoo.com> wrote:
> I am just trying to learn Mathematica. What am I doing wrong here?
>
> I have a very simple equation:
>
> x^3 - 15 x + 2 = 0
>
> When I plot it in Mathematica 7,
>
> ClearAll[*]
> f[x_] := x^3 - 15 x + 2
> Plot[f[x], {x, -5, 5}]
>
>
> it gives the expected graph of a cubic, with three real roots near -4,
> 0, and 4.
>
>
> When I NSolve it,
>
> NSolve[f[x] == 0, x]
>
> it gives
>
> {{x -> -3.938}, {x -> 0.133492}, {x -> 3.80451}}
>
> which is exactly what you would expect from the graph.
>
> But when I Solve it
>
> Solve[f[x] == 0, x]
>
> it gives this mess
>
> {{x -> 5/(-1 + 2 I Sqrt[31])^(1/3) + (-1 + 2 I Sqrt[31])^(
> 1/3)}, {x -> -((5 (1 + I Sqrt[3]))/(
> 2 (-1 + 2 I Sqrt[31])^(1/3))) -
> 1/2 (1 - I Sqrt[3]) (-1 + 2 I Sqrt[31])^(1/3)}, {x -> -((
> 5 (1 - I Sqrt[3]))/(2 (-1 + 2 I Sqrt[31])^(1/3))) -
> 1/2 (1 + I Sqrt[3]) (-1 + 2 I Sqrt[31])^(1/3)}}
>
>
> I don't know how it looks in your font, but that "I" in each solution
> is the imaginary i. Solve is saying this equation has no real roots,
> even though the graph clearly shows that all three roots are real.
>
> Can someone tell me if I am doing something wrong, or am I expecting
> something wrong, or if I just can't trust Mathematica? Thanks for any
> help.
>
--
DrMajorBob at yahoo.com
- References:
- Bug in Solve?
- From: tonysin <a2mgoog@yahoo.com>
- Bug in Solve?