22 Nov
2010
22 Nov
'10
11:32 a.m.
On 11/22/10, Thomas Colthurst <thomaswc@gmail.com> wrote:
I don't know the answer to Mike's question, but I do have two observations about the roots of p'(x) that may be of interest. ... One consequence of this is that for any point z and multiset of n-1 points Y, there is another multiset of n-1 points Z such that C(z union Z) = Y. Just coordinate shift so that z is the origin, then integrate p'(x) to get p(x) + k. But p(0) = 0, so k = 0.
"multiset of n points Y", I think ... ? Interesting points you made there, TC. WFL