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

*To*: mathgroup at smc.vnet.net*Subject*: [mg128499] Re: How accurate is the solution for high degree algebraic equation?*From*: Sseziwa Mukasa <mukasa at gmail.com>*Date*: Thu, 25 Oct 2012 01:43:14 -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>

For a polynomial of such large degree it's unlikely to can use machine precision. Mathematica can solve this in infinite precision quickly: (Debug) In[12]:= d = 54; f = (-z - 1)^d - (-z^d - 1); sol = Solve[f == 0, z]; a = z /. sol; (Debug) In[17]:= FullSimplify[f /. z -> a[[1]]] (Debug) Out[17]= 0 You can see numerical approximations of the solutions using N: (Debug) In[20]:= N[a[[1]]] N[a[[1]], 30] (Debug) Out[20]= -0.5 - 17.1839 I (Debug) Out[21]= -0.5000000000000000000000000000 - 17.1838854436050918792404513804 I On Oct 24, 2012, at 3:32 AM, Alexandra 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>