Re: Why is the negative root?

• To: mathgroup at smc.vnet.net
• Subject: [mg69481] Re: Why is the negative root?
• From: dimmechan at yahoo.com
• Date: Thu, 14 Sep 2006 06:54:49 -0400 (EDT)
• References: <ee8h8v\$j48\$1@smc.vnet.net>

```***(I have converted everything to InputForm)
***Indeed, you have right

(Reduce[{z^3-z^2- b z-1==0, b>0, z>0}, z] //ToRadicals)/.b->3.//Chop
z == -0.4142135623730952

***However, you want the positive solution:

Select[z/.{ToRules[Reduce[z^3-z^2- b z-1==0,z]/.b->3]},Positive]//N
{2.414213562373095}

*** So, let's type your commands step by step to see what happens

Reduce[{z^3-z^2- b z-1==0, b>0, z>0}, z]
(Inequality[0, Less, b, Less, Root[-31 - 18*#1 + #1^2 + 4*#1^3 & , 1,
0]] && z == Root[-1 - b*#1 - #1^2 + #1^3 & , 1]) ||
(b >= Root[-31 - 18*#1 + #1^2 + 4*#1^3 & , 1, 0] && z == Root[-1 -
b*#1 - #1^2 + #1^3 & , 3])

(Inequality[0, Less, b, Less, -1/12 + (6371 - 624*Sqrt[78])^(1/3)/12 +
(6371 + 624*Sqrt[78])^(1/3)/12] &&
z == 1/3 - (2^(1/3)*(-1 - 3*b))/(3*(29 + 9*b + 3*Sqrt[3]*Sqrt[31 +
18*b - b^2 - 4*b^3])^(1/3)) +
(29 + 9*b + 3*Sqrt[3]*Sqrt[31 + 18*b - b^2 -
4*b^3])^(1/3)/(3*2^(1/3))) ||
(b >= -1/12 + (6371 - 624*Sqrt[78])^(1/3)/12 + (6371 +
624*Sqrt[78])^(1/3)/12 &&
z == 1/3 + ((1 - I*Sqrt[3])*(-1 - 3*b))/(3*2^(2/3)*(29 + 9*b +
3*Sqrt[3]*Sqrt[31 + 18*b - b^2 - 4*b^3])^(1/3)) -
((1 + I*Sqrt[3])*(29 + 9*b + 3*Sqrt[3]*Sqrt[31 + 18*b - b^2 -
4*b^3])^(1/3))/(6*2^(1/3)))

***Now it will be very helpful to execute the N command

N[%]
(Inequality[0., Less, b, Less, 2.6107186132760383] &&
z == 0.3333333333333333 - (0.41997368329829105*(-1. - 3.*b))/
(29. + 9.*b + 5.196152422706632*Sqrt[31. + 18.*b - 1.*b^2 -
4.*b^3])^(1/3) +
0.26456684199469993*(29. + 9.*b + 5.196152422706632*Sqrt[31. +
18.*b - 1.*b^2 - 4.*b^3])^(1/3)) ||
(b >= 2.6107186132760383 && z == 0.3333333333333333 +
((0.20998684164914555 - 0.3637078786572405*I)*(-1. - 3.*b))/
(29. + 9.*b + 5.196152422706632*Sqrt[31. + 18.*b - 1.*b^2 -
4.*b^3])^(1/3) - (0.13228342099734997 + 0.22912160616643376*I)*
(29. + 9.*b + 5.196152422706632*Sqrt[31. + 18.*b - 1.*b^2 -
4.*b^3])^(1/3))

***See now what happens:

%/.b->3//Chop
z == -0.41421356237309515

***I.e. b->3 satisfies the second inequality (b >= 2.6107186132760383 )
which is respectful to the negative root!

***Executing the following command gives what you want:

(Reduce[{z^3-z^2- b z-1==0, b>0, z>0}, z]/.b->3//ToRadicals)//N
z == 2.414213562373095

Cheers,
Dimitris

Î?/Î? p-valko at tamu.edu Î­Î³Ï?Î±Ï?Îµ:
> (Reduce[{z^3-z^2- b z-1==0, b>0, z>0}, z] //ToRadicals)/.b->3.//Chop
>
> gives a negativ z:
>
> z == - 0.414214
>
> What am I doing wrong?
>
> Regards
> Peter

```

• Prev by Date: Re: Re: Re: solve and Abs
• Next by Date: Re: Locating common subexpressions
• Previous by thread: Re: Why is the negative root?
• Next by thread: Re: Re: Why is the negative root?