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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
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*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?
>
>
>
>
>
>
>
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