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how can one prove that there exists a sufficiently small $\epsilon>0$ s.t. the polynomial $\sum\limits_{i=0}^m a_i \epsilon^i \ge 0$, where there is an index $k$ such that $a_k=0$ for $a0$. Any estimation tips or other tips?

Thanks.

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    In what sense if $a$ ordered, i. e. what inequalities do hold? And what is $u_k$? – 2017-01-07
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    I find your natation very cunfusing. $a_k$ are your coefficients, which are (assumed) real, where $k$ is an (integer) index. You can't compare them and get something logcal as result. In addition, we now, that $\begin{pmatrix}a_0, a_1, ..., a_m \end{pmatrix}$ is lexicographilly ordered, since that is the order, you wrote thmen down. This additionally means, that there are now constraints on their positivy... – 2017-01-07
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    @martini: "lexicographically order" is a misuse of the term by the OP. The meaning, though, is given in the post: there exist a $k$ such that the coefficients of the polynomial are $(0, \dots, 0, a_k, a_{k+1}, \dots, a_m)$ with $a_k > 0$. Un unhappy choice of words by the OP, I agree. – 2017-01-07
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    Sorry, i meant lexicographically positive. – 2017-01-07

1 Answers 1

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Your polynomial is

$$f = a_m x^m + a_{m-1} x^{m-1} + \dots + a_k x^k$$

with $a_k > 0$. Let

$$g = a_m x^{m-k} + a_{m-1} x^{m-k-1} + \dots + a_k .$$

Notice that $g$ is continuous and that $g(0) = a_k > 0$. By continuity, it follows that there exist a $r > 0$ such that for $|x| < r$ we have $g(x) > 0$. Pick any $\epsilon \in (0,r)$. Then $f(\epsilon) = \epsilon^k g(\epsilon) > 0$.