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 times greater, so the polynomial is now amazingly
ill-conditioned. Note that if you evaluate the translated polynomial in
machine floating-point, and then translate back, you get coefficients of
size 10^15. This translation is not feasible in floating-point!
On Saturday, 14 December 2013 17:37:16 UTC, JamesHDavenport wrote:
>
> Curious. The polynomial does not seem particularly ill-conditioned, in the
> sense that its discriminant 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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