Sorry - I was looking at the original *(I*x^51+sum(x^k,k,0,50))==0.* Note
that if you make the x->x-1 substitution, the polynomial now has
coefficient as large as 10^18. The discriminant is unchanged, but the value
you would expect it to be, given the size of the coefficient, is
roughly10^800 t
Apparently you need about 65 bits of accuracy, presumably maxima uses
doubles (=53 bits). Compare:
sage: eq = (I*x^51+sum(x^k,k,0,50)).subs(x=x-1-I)
sage: var('x, k')
(x, k)
sage: eq = (I*x^51+sum(x^k,k,0,50)).subs(x=x-1-I)
sage: sol = eq.polynomial(ComplexField(65)).roots()
sage: [abs(eq.subs(x=
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
