Let $V(\text{sequence})$ be the number of sign changes in the sequence, e.g. $V(-3,0,-2,9,0,1)=1$. Show that $V(a_0,a_1,...,a_n)\ge V(a_0,a_0+a_1,a_0+a_1+a_2,...)$. Furthermore, prove that if $\sum_{i=0}^na_i=0$, then the number of positive roots of $p(x)=\sum_{i=0}^na_ix^i$ is at most $V(a_0,a_0+a_1,a_0+a_1+a_2,...)+1$.
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1How is the "number of sign changes" in your example 1? I count 3. Is it the product of the signs of the sequence elements instead? – gnometorule Jun 28 '14 at 13:58
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Zeroes are ignored. So the only change of sign is -2 -> 9. – Guest123 Jun 28 '14 at 14:05
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Since $p(1)=0$, perform polynomial division of $p(x)$ by $x-1$ and examine the resulting coefficient sequence. It should look familiar.
Lutz Lehmann
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