Re: How accurate is the solution for high degree algebraic
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- Subject: [mg128490] Re: How accurate is the solution for high degree algebraic
- From: Fred Simons <f.h.simons at tue.nl>
- Date: Thu, 25 Oct 2012 01:40:12 -0400 (EDT)
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To answer your question: no, the degree is not too high, but you are
using the wrong number system, machine numbers. The polynomial has a
high degree, so the derivative has a high degree as well, which means
that a small change in an argument will give a tremendous change in the
function value. Machine numbers are poorly suited to deal with this
Mathematica can solve your equation exact, using root expressions:
Let us look at the first solution:
In:= f /. sol[] // RootReduce
So it IS a solution. Let us see what happens if we use machine numbers:
In:= f /. N[sol[]]
Out= -3.82475*10^50 + 0. I
When we use arbitrary precision numbers, you get a more reliable result:
In:= f /. N[sol[], 100]
Out= 0.*10^-31 + 0.*10^-31 I
Indeed this is close to zero. Observe that the precision went down
almost 70 digits!
Eindhoven University of Technology
Op 24-10-2012 9:32, Alexandra schreef:
> 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[] // N
> It returns a number very close to zero. This is natural.
> But if I compute
> f /. (z -> a[]) // N
> Mathematica returns
> 12.0047 + 14.7528 I
> I cannot say a[] 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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