# Positive polynomial conditions

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 $$a_N>0$$?

• Why the RHS lost its $a_i$s? Jun 24, 2019 at 11:18
• @user10354138 sry, edited.
– anon
Jun 24, 2019 at 11:21
• There is no real need for the denominators $i!$, you can very well work with the coefficients $b_i=a_i/i!$.
– user65203
Jun 24, 2019 at 11:46
• @YvesDaoust Yes, indeed. Edited.
– anon
Jun 24, 2019 at 11:50
• 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. Jun 24, 2019 at 11:51

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 $$x$$.
• so, basically, the polynomial has to have no roots on $(0, +\infty)$, and $a_N>0$
• $p_0 = p = \sum_{i=0}^n a_i x^i$, $p_1 = p' = \sum_{i=1}^n i a_i x^{i-1}$, $p_{i+1} = - \text{rem}(p_{i-1}, p_i)$: its coefficients are rational functions of the coefficients of $p_i$ and $p_{i+1}$, but I don't know of a general expression. Jun 24, 2019 at 17:52