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Suppose we have a polynomial $p$

$$ p = a_0 + a_1x + a_2x^2 + \dots+a_Nx^N = \sum_{i=0}^Na_ix^i. $$

what is the sufficient and necessary condition of $\{a_i:i=0,1,\dots, N\}$ such that $p$ is always positive for all $x\in(0, +\infty)$

(does such sufficient and necessary conditions even exist?)

In other words: what kind of (and necessary) coefficients could make the polynomial always positive on the right side of $\mathbb{R}?$

Guess: what about let $\xi_k$ be all of the roots of $\frac{dp}{dx}=0$, and the condition is $p(\xi_k)$ are all positive, and $a_N>0$?

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  • $\begingroup$ Why the RHS lost its $a_i$s? $\endgroup$
    – user10354138
    Jun 24 '19 at 11:18
  • $\begingroup$ @user10354138 sry, edited. $\endgroup$
    – Nathan Explosion
    Jun 24 '19 at 11:21
  • $\begingroup$ There is no real need for the denominators $i!$, you can very well work with the coefficients $b_i=a_i/i!$. $\endgroup$
    – user65203
    Jun 24 '19 at 11:46
  • $\begingroup$ @YvesDaoust Yes, indeed. Edited. $\endgroup$
    – Nathan Explosion
    Jun 24 '19 at 11:50
  • $\begingroup$ Your guess involved roost and it is not completely in terms of the coefficients. In terms of roots a N and S condition is that all the roots are negative. But I don't know if this can be written in terms of the coefficients. $\endgroup$
    – Kavi Rama Murthy
    Jun 24 '19 at 11:51
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A necessary and sufficient condition uses Sturm's theorem. You want $a_N > 0$ and $V(0) = V(\infty)$, where $V(x)$ is the number of sign changes of the Sturm sequence of $p$ at $x$.

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  • $\begingroup$ so, basically, the polynomial has to have no roots on $(0, +\infty)$, and $a_N>0$ $\endgroup$
    – Nathan Explosion
    Jun 24 '19 at 12:46
  • $\begingroup$ May I further ask, if using the polynomial in my question, are there any general expression for the Sturm's sequence in terms of those coefficients? $\endgroup$
    – Nathan Explosion
    Jun 24 '19 at 15:19
  • $\begingroup$ $p_0 = p = \sum_{i=0}^n a_i x^i$, $p_1 = p' = \sum_{i=1}^n i a_i x^{i-1}$, $p_{i+1} = - \text{rem}(p_{i-1}, p_i)$: its coefficients are rational functions of the coefficients of $p_i$ and $p_{i+1}$, but I don't know of a general expression. $\endgroup$
    – Robert Israel
    Jun 24 '19 at 17:52

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