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I encountered several polynomials as below:

$$f(x)=7 + 91 x - 385 x^2 + 1659 x^3 - 1379 x^4 + 553 x^5 - 35 x^6 + x^7$$ $$g(x)=33 + 110 x + 495 x^2 - 252 x^3 + 335 x^4 - 18 x^5 + x^6$$ $$h(x)=71 + 237 x + 126 x^2 + 210 x^3 - 5 x^4 + x^5$$

When I plot them from $x=0$ to $x=10$, they seemed to be positive, i.e., they do not have positive roots.

For these numerical samples, one could use Sturm sequence to determine the number of positive roots in the interval $(0,\infty)$. But we hesitate to use it because for arbitrary real coefficients $a_k$, we will encounter divisions.

Question: for a polynomial with real coefficients like $p_n(x)=\sum_{k=0}^{n}a_k x^k$, are there any simple and sufficient conditions like $$ c_k a_{k-1}a_{k+1}<a_k^2<b_k a_{k-1}a_{k+1},$$ where $b_k, c_k$ are know functions of $k$, such that $p_n(x)$ does not have any positive roots?

mike
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  • @TravisJ Thanks for the comment. I modified my question to delete the positive requirement for $a_k$. – mike Jul 02 '15 at 12:45
  • Yes. $x=-0.0585666$ is a root. All the other roots are nonreal and having positive real part. – mike Jul 02 '15 at 12:56

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Here is an old result by Poincare (references can be found in the AMS Memoir by David Handelman, Positive polynomials and product type actions of compact groups, Memoirs AMS 320):

A polynomial $p(x)$ with $p(0)>0$ has no positive roots if and only if there is some $n$ such that $(1+x)^n p(x)$ has only positive coefficients. One can derive conditions on the coefficients of $p(x)$ by expanding the binomial $(1+x)^n$, but it is in general hard to know what $n$ will work, and there are examples that show that $n$ may need to be arbitrarily large, even for polynomials with only 5 terms.

For example, for the polynomial $f(x)$ given in the question, $(1+x)^{48} f(x)$ has only positive coefficients, so $f(x)$ has no positive roots.

Valerio
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