Square Roots and Sequences
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From: G. A. Edgar on 15 Apr 2010 08:07 In article <938b684b-26ba-44b5-b5bf-5ad644e28223(a)h27g2000yqm.googlegroups.com>, Ludovicus <luiroto(a)yahoo.com> wrote: > On Apr 14, 9:45�am, Maury Barbato <mauriziobarb...(a)aruba.it> wrote: > > > > So {t_n} is convergent, and the limit t is the unique > > solution of the equation t^2 = 2 + t^(1/2). > > The solution of the equation: t = [2 + t^(1/2)]^(1/2) > is 1.83117720722624... > I don't know if it is possible to express it algebraically. > > According to Maple, ((1/6)*(188+12*249^(1/2))^(1/3)-(4/3)/(188+12*249^(1/2))^(1/3)+1/3)^2 -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/
From: Arturo Magidin on 15 Apr 2010 12:08 On Apr 15, 3:18 am, Ludovicus <luir...(a)yahoo.com> wrote: > On Apr 14, 9:45 am, Maury Barbato <mauriziobarb...(a)aruba.it> wrote: > > > So {t_n} is convergent, and the limit t is the unique > > solution of the equation t^2 = 2 + t^(1/2). > > The solution of the equation: t = [2 + t^(1/2)]^(1/2) > is 1.83117720722624... > I don't know if it is possible to express it algebraically. Since t^2 = 2+t^{1/2} is a quartic equation on t^{1/2}, it is solvable by radicals. A bit complicated, but it can certainly be expressed algebraically. Of course, only the solution that is real, positive and in the correct region will matter. -- Arturo Magidin
From: Chip Eastham on 15 Apr 2010 15:35 On Apr 15, 12:08 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > On Apr 15, 3:18 am, Ludovicus <luir...(a)yahoo.com> wrote: > > > On Apr 14, 9:45 am, Maury Barbato <mauriziobarb...(a)aruba.it> wrote: > > > > So {t_n} is convergent, and the limit t is the unique > > > solution of the equation t^2 = 2 + t^(1/2). > > > The solution of the equation: t = [2 + t^(1/2)]^(1/2) > > is 1.83117720722624... > > I don't know if it is possible to express it algebraically. > > Since t^2 = 2+t^{1/2} is a quartic equation on t^{1/2}, it is solvable > by radicals. A bit complicated, but it can certainly be expressed > algebraically. Of course, only the solution that is real, positive and > in the correct region will matter. Hi, Arturo: Yes, in fact the quartic has a rational root t^(1/2) = -1 that is easily "deflated", so we only need to solve a cubic (which turns out to have the one positive real root and a complex conjugate pair of roots): r^3 - r^2 + r - 2 = 0 regards, chip
From: Rob Johnson on 15 Apr 2010 16:31
In article <3a654ed0-a5af-44e3-9c3d-0ccea1b5cb7e(a)r1g2000yqb.googlegroups.com>, Arturo Magidin <magidin(a)member.ams.org> wrote: >On Apr 15, 3:18 am, Ludovicus <luir...(a)yahoo.com> wrote: >> On Apr 14, 9:45 am, Maury Barbato <mauriziobarb...(a)aruba.it> wrote: >> >> > So {t_n} is convergent, and the limit t is the unique >> > solution of the equation t^2 = 2 + t^(1/2). >> >> The solution of the equation: t = [2 + t^(1/2)]^(1/2) >> is 1.83117720722624... >> I don't know if it is possible to express it algebraically. > >Since t^2 = 2+t^{1/2} is a quartic equation on t^{1/2}, it is solvable >by radicals. A bit complicated, but it can certainly be expressed >algebraically. Of course, only the solution that is real, positive and >in the correct region will matter. The quartic equation for t is t^4 - 4 t^2 - t + 4 = 0 This has a trivial factor of t - 1, but t = 1 does not satisfy the original equation. Factoring this solution out gives t^3 + t^2 - 3 t - 4 = 0 This equation has only one real solution. Mathematica gives -1/3 + (316 - 12 Sqrt[249])^(1/3)/6 + (316 + 12 Sqrt[249])^(1/3)/6 Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font |