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

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  • 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)
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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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