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Topic: How accurate is the solution for high degree algebraic equation?
Replies: 5   Last Post: Oct 25, 2012 11:39 PM

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Simons, F.H.

Posts: 107
Registered: 12/7/04
Re: How accurate is the solution for high degree algebraic
Posted: Oct 25, 2012 1:44 AM
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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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