Timeline for Positive roots of a polynomial

10 events

when toggle format	what		by	license	comment
May 4, 2015 at 12:59	vote	accept	dima
May 4, 2015 at 12:30	answer	added	Max Alekseyev		timeline score: 21
May 4, 2015 at 4:25	comment	added	Suvrit		For $n=3$ and also for $n=4$ it seems there is just one change of sign in the consecutive terms, so Descartes' rule of signs should kick in and yield at most one positive root. Perhaps one can even inductively show that there is just one change of sign, and a small additional argument shows that there is at least one positive root.
May 3, 2015 at 20:07	comment	added	Douglas Zare		If you divide by $\prod (x+a_k)^2$, you can replace the product with the reciprocal of the omitted term, $(x+a_i)^{-2}$.
May 3, 2015 at 19:51	comment	added	dima		@Greg Martin: $n=1$ does not satisfy the condition that not all $a_i$'s are equal. For $n=2$ we get $p(x)=\frac{1}{2} \left(a_1-a_2\right){}^2 x^2-\frac{1}{2} a_1 \left(a_1-a_2\right){}^2 a_2$ and so clearly the only positive root is $x_0=\sqrt{a_1 a_2}$. For $n=3$ I don't know already...
May 3, 2015 at 19:21	comment	added	Greg Martin		Can you prove it for $n=1$, $n=2$, $n=3$?
May 3, 2015 at 9:00	history	edited	dima	CC BY-SA 3.0	added 55 characters in body
May 3, 2015 at 8:59	comment	added	dima		@Seva some numerical evidence, I'll add this to the question
May 3, 2015 at 8:58	comment	added	Seva		What makes you tho expect that it has exactly one positive root?
May 3, 2015 at 8:23	history	asked	dima	CC BY-SA 3.0