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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

 Messages: [ Previous | Next ]
 Sseziwa Mukasa Posts: 108 Registered: 8/26/07
Re: How accurate is the solution for high degree algebraic equation?
Posted: Oct 25, 2012 1:41 AM

For a polynomial of such large degree it's unlikely to can use machine precision. Mathematica can solve this in infinite precision quickly:

(Debug) In[12]:= d = 54; f = (-z - 1)^d - (-z^d - 1);
sol = Solve[f == 0, z];
a = z /. sol;
(Debug) In[17]:= FullSimplify[f /. z -> a[[1]]]
(Debug) Out[17]= 0

You can see numerical approximations of the solutions using N:

(Debug) In[20]:= N[a[[1]]]
N[a[[1]], 30]
(Debug) Out[20]= -0.5 - 17.1839 I
(Debug) Out[21]= -0.5000000000000000000000000000 - 17.1838854436050918792404513804 I

On Oct 24, 2012, at 3:32 AM, Alexandra 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?
>
>
>
>
>
>
>

Date Subject Author
10/25/12 Sseziwa Mukasa
10/25/12 Bob Hanlon
10/25/12 Simons, F.H.
10/25/12 Alexei Boulbitch
10/25/12 Andrzej Kozlowski