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]) ToRadicals[%] (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 ÎγÏ?αÏ?ε: > (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