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Re: How accurate is the solution for high degree algebraic equation?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg128499] Re: How accurate is the solution for high degree algebraic equation?
*From*: Sseziwa Mukasa <mukasa at gmail.com>
*Date*: Thu, 25 Oct 2012 01:43:14 -0400 (EDT)
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*References*: <20121024073227.62DD76885@smc.vnet.net>
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?
>
>
>
>
>
>
>
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