[newbie question] solving fifth order polynomial equation
Subject: [newbie question] solving fifth order polynomial equation
From: Hugo Coolens
Date: Wed, 17 Jan 2007 19:23:58 +0100 (CET)
Thanks for the helpful and interesting reply, I just have one little
problem mentioned below:
On Wed, 17 Jan 2007, Stavros Macrakis wrote:
> On 1/17/07, Hugo Coolens <coolens at kahosl.be> wrote:
>> When solving the following equation
>> solve(1+11*s+45*s^2+84*s^3+70*s^4+21*s^5=0);
>>
>> Maxima shows two real solutions and three complex ones in stead of five
>> real solutions.
>
> Actually, Maxima is showing five real solutions -- it's just that the
> *form* of some of the real solutions includes %i. To simplify these
> expressions to purely real expressions, use radcan:
>
> radcan(solve(1+11*s+45*s^2+84*s^3+70*s^4+21*s^5=0)) =>
> [s = -1/3,s = -1,
> s = -(sqrt(3)*sin((atan(9/(13*sqrt(3)))-%pi)/3)
> +cos((atan(9/(13*sqrt(3)))-%pi)/3)+2)
> /3,
> s = (sqrt(3)*sin((atan(9/(13*sqrt(3)))-%pi)/3)
> -cos((atan(9/(13*sqrt(3)))-%pi)/3)-2)
> /3,s = (2*cos((atan(9/(13*sqrt(3)))-%pi)/3)-2)/3]
Strange enough my system does not give me the "real" values as you can see
here, any idea why?:
radcan(solve(1+11*s+45*s^2+84*s^3+70*s^4+21*s^5=0));
(%o10) [s = -1/3,s = -1,
s = -((9*%I-13*SQRT(3))^(2/3)*(SQRT(3)*%I+1)
+4*2^(1/3)*3^(1/6)*7^(1/3)*(9*%I-13*SQRT(3))^(1/3)
-2^(2/3)*3^(5/6)*7^(2/3)*%I+2^(2/3)*3^(1/3)*7^(2/3))
/(6*2^(1/3)*3^(1/6)*7^(1/3)*(9*%I-13*SQRT(3))^(1/3)),
s = ((9*%I-13*SQRT(3))^(2/3)*(SQRT(3)*%I-1)
-4*2^(1/3)*3^(1/6)*7^(1/3)*(9*%I-13*SQRT(3))^(1/3)
-2^(2/3)*3^(5/6)*7^(2/3)*%I-2^(2/3)*3^(1/3)*7^(2/3))
/(6*2^(1/3)*3^(1/6)*7^(1/3)*(9*%I-13*SQRT(3))^(1/3)),
s = ((9*%I-13*SQRT(3))^(2/3)-2*2^(1/3)*3^(1/6)*7^(1/3)
*(9*%I-13*SQRT(3))^(1/3)
+2^(2/3)*3^(1/3)*7^(2/3))
/(3*2^(1/3)*3^(1/6)*7^(1/3)*(9*%I-13*SQRT(3))^(1/3))]