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Re: How accurate is the solution for high degree algebraic
*To*: mathgroup at smc.vnet.net
*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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Alexandra,
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
behaviour.
Mathematica can solve your equation exact, using root expressions:
d=54;f=(-z-1)^d-(-z^d-1);
sol=Solve[f==0,z];
Let us look at the first solution:
In[4]:= f /. sol[[1]] // RootReduce
Out[4]= 0
So it IS a solution. Let us see what happens if we use machine numbers:
In[5]:= f /. N[sol[[1]]]
Out[5]= -3.82475*10^50 + 0. I
When we use arbitrary precision numbers, you get a more reliable result:
In[6]:= f /. N[sol[[1]], 100]
Out[6]= 0.*10^-31 + 0.*10^-31 I
Indeed this is close to zero. Observe that the precision went down
almost 70 digits!
Regards,
Fred Simons
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[[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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