Smallest root of a set of polynomials

Solution 1:

Let $f(x) = \sum\limits_{k=0}^m c_k x^k \in \mathcal{P}_{0,1}$ be any polynomial whose most negative real root $\alpha < -1$.

If $c_0 = 0$, we can factor $f(x)$ as $x^n g(x)$ for some $n > 0$ and $g(x) \in \mathcal{P}_{0,1}$ which satisfies $g(0) = 1$.
Since $g(x)$ also has $\alpha$ as its most negative real root, WOLOG, we will assume $c_0 = c_m = 1$.

Let $\beta = -\frac{1}{\alpha} \in (0,1)$. Consider the polynomial $h(x) \stackrel{def}{=} x^m f(\frac1x) = \sum_{k=0}^m c_{m-k} x^k$. It also $\in \mathcal{P}_{0,1}$ and having $-\beta$ as its most positive negative real root. Decompose $h(x)$ as $1 + xp(x^2) + q(x^2)$ where $p(x), q(x) \in \mathcal{P}_{0,1}$, we have

$$h(-\beta) = 1 - \beta p(\beta^2) + q(\beta^2) = 0 \quad\implies\quad \beta p(\beta^2) \ge 1$$ Since $p(\beta^2) < \sum_{k=0}^\infty \beta^{2k} = \frac{1}{1-\beta^2}$, this leads to $$\frac{\beta}{1-\beta^2} > 1 \quad\iff\quad \beta > \frac{1}{\phi} \quad\implies\quad \alpha > -\phi $$ As a result, we have

$$t \stackrel{def}{=} \inf\{\alpha \in \mathbb{R} : \exists P \in \mathcal{P}_{0,1}, \, P(\alpha)=0\} \ge - \phi\tag{*1}$$

For any $n > 1$, consider the polynomial

$$f_n(x) = 1 + x^{2n} + x \sum_{k=0}^{n-1} x^2 = 1+x^{2n} + \frac{x}{x^2-1} (x^{2n}-1)$$ Since $\frac{\phi}{\phi^2-1} = 1$, we have $f_n(-\phi) = 2$. Notice

$$f'_n(x) = 2n x^{2n-1}\left(1 + \frac{x}{x^2-1}\right) + \left(\frac{1}{x^2-1} - \frac{2x^2}{(x^2-1)^2}\right) (x^{2n} - 1 )$$ We have $f'_n(-\phi) = -(3-\phi)(\phi^{2n}-1)$.

Since this is negative and glowing exponentially in magnitude as $n \to \infty$, $f_n(x)$ has a negative root $\alpha_n$ near $-\phi$ which satisfies:

$$\alpha_n + \phi \approx \frac{2}{(3-\phi)(\phi^{2n}-1)} \to 0\quad\text{ as }\quad n \to \infty$$

This implies

$$t \le \lim_{n\to\infty} \alpha_n = -\phi\tag{*2}$$

Combine $(*1)$ and $(*2)$, we have $t = -\phi$.