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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? > > > > > > >



