Re: Bug in Solve?
- To: mathgroup at smc.vnet.net
- Subject: [mg102949] Re: Bug in Solve?
- From: Albert Retey <awnl at gmx-topmail.de>
- Date: Thu, 3 Sep 2009 05:37:12 -0400 (EDT)
- References: <h7l8rh$35s$1@smc.vnet.net>
tonysin 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.
Nothing wrong, you just need to realize that an expression can be real
valued, even if it contains I at some point, e.g.: Exp[2 Pi I]
It is easy to check that this is the case here:
Solve[f[x] == 0, x]//N
actually you will find that there will be left some very small imaginary
parts, which you could get rid off with:
Solve[f[x] == 0, x] // N // Chop
If you don't trust the numerics, you can also do this to check that the
imaginary parts are indeed zero for all three solutions:
FullSimplify[Im[x] /. Solve[x^3 - 15 x + 2 == 0, x]]
hth,
albert