Curious. The polynomial does not seem particularly ill-conditioned, in the
sense that its discriminat is roughly what one might expect (unlike, say,
Wilkinson's). Maple gives 50 roots with |f(x)|<10^{-12}, and one with f(x)=,
i.e. much larger.
On Thursday, 12 December 2013 15:35:53 UTC, AWWQUB wrote:
>
> Consider the equation
> *(I*x^51+sum(x^k,k,0,50))==0*
> Try to solve it numerically using
> *solve([(I*x^51+sum(x^k,k,0,50))==0,x==x],x,solution_dict=True)*
> and you obtain 51 solutions of which 50 have modulus approximately 1 and
> the other is close to 1+I. Substituting back gives residuals of around at
> most 10^-6. That looks fine and Mathematica gives similar solutions. Now
> substitute x-1-I for x, so that one of the solutions should now be close to
> zero. Sage now gives solutions with real and imaginary parts both between
> -1 and +4, but none are particularly close to zero.The residuals now range
> up to 10^20. Mathematica gives completely different answers but which are
> also wrong, namely all 51 are approximately 1.
>
> Any ideas?
>
> Tony Wickstead
>
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