Re: How accurate is the solution for high degree algebraic equation?

*To*: mathgroup at smc.vnet.net*Subject*: [mg128501] Re: How accurate is the solution for high degree algebraic equation?*From*: Bob Hanlon <hanlonr357 at gmail.com>*Date*: Thu, 25 Oct 2012 01:43:55 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net*References*: <20121024073227.62DD76885@smc.vnet.net>

Use higher precision calculations d = 54; f[z_] = (-z - 1)^d - (-z^d - 1); sol = Solve[f[z] == 0, z]; Length[sol] 54 With only machine precision f[z] /. sol[[3]] // N -8.4784*10^25 + 0. I With 30 digit precision f[z] /. sol[[3]] // N[#, 30] & 0.*10^-37 + 0.*10^-37 I % // Chop 0 Off[N::meprec] f[z] /. sol // N[#, 30] & // Chop // Union {0} Bob Hanlon On Wed, Oct 24, 2012 at 3:32 AM, Alexandra <watanabe.junzo at gmail.com> wrote: > I wanted to know all the solutions of f = (-z - 1)^d - (-z^d - 1)==0, where d=54. > I did the following: > > d = 54; f = (-z - 1)^d - (-z^d - 1); > sol = NSolve[f == 0,z]; > a = z /. sol; > > So a is a set of solutions. > > If I compute > f /. z -> a[[50]] // N > It returns a number very close to zero. This is natural. > > But if I compute > f /. (z -> a[[1]]) // N > > Then > Mathematica returns > 12.0047 + 14.7528 I > > I cannot say a[[1]] is a solution of f=0. > > Many other elements in the solution set a does not seem to satisfy the equation. > Only the last few terms in a are satisfactory enough as solutions. > > Is the degree too high? > > > > > > >

**References**:**How accurate is the solution for high degree algebraic equation?***From:*Alexandra <watanabe.junzo@gmail.com>