Conjecture:
If A is a nonempty, finite set with size |A|=n, and
PA(x)=∏a∈A(x−a),
then PA(x) can be expanded as
PA(x)=n∑k=0(−1)n−kxk∑U⊆A|U|=n−k∏u∈Uu.
I have conjectured this based on algebraic evidence. That is, I expanded out the cases n=1,...,4 both manually and through (1), and in each case the conjecture held. The problem is, I'm having difficulty proving this result. I'm fairly certain that a proof would involve induction, but I'm not very good at that, and have so far failed.
I was initially interested in this formula because I recognized that (1) would imply that
k≠0,n⟺∑U⊆Sn|U|=k∏u∈Uu=0
for
Sn={exp2πikn:k=0,1,...,n−1}.
It may seem un-intuitive at first, but (2) is true because
PSn(x)=xn−1,
(as Sn is the set of roots of xn−1) so each term of the expansion must vanish except for the cases k=0,n.
After seeing this, I was naturally curious about a proof of (1). Could I have some help or hints? Thanks.
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