I'm looking for a proof of the theorem 7, page 6, of this document :


Theorem 7 (E. Laguerre) Let $f \in \mathbb{R}[x]$ be a polynomial of degree $n \ge 2$, that has only real simple roots and that satisfies the second–order differential equation $$ p(x)y'' + q(x)y' + r(x) y = 0,$$ with $p$, $q$ and $r$ univariate polynomials with real coefficients, $p(x) \ne 0$. If $\alpha$ is a root of the polynomial $f$, then we have $$4(n−1)\left( p(\alpha)r(\alpha)+p(\alpha)q'(\alpha) - p'(\alpha)q(\alpha)\right) - (n+2) q(\alpha)^2 \ge 0$$

I don't find any proof on the internet and I failed to prove it myself. Does anyone know ?



WOLOG, let us assume $\alpha = 0$ and I'll only consider the case $n > 2$. To shorten expressions, we will use a suffix $0$ to denote the value of a polynomial evaluate at $0$.

Differentiate the ODE

$$ p(x)y'' + q(x)y' + r(x) y = 0,\tag{*1}$$

one more times and using the assumption $y_0 \stackrel{def}{=} y(0) = 0$, it is easy to show

$$\begin{cases} y_0'' &= -\frac{q_0}{p_0} y_0'\\ y_0''' &= -\frac{1}{p_0}\left( (q_0 + p_0')\left(-\frac{q_0}{p_0}\right) + (r_0 + q_0')\right) y_0' \end{cases}\tag{*2}$$

Let $\lambda_1, \lambda_2, \cdots, \lambda_{n-1}$ be the non-zero roots of $y(x)$. Consider the polynomial

$$g(x) = \prod_{k=1}^{n-1} (x - \lambda_k) = x^{n-1} - \sigma_1 x^{n-2} + \cdots + (-1)^{n-2} \sigma_{n-2} x + (-1)^{n-1} \sigma_{n-1}$$

Up to a sign, the coefficients of $g(x)$ is the $k^{th}$ elementary symmetric polynomial for the list of $n-1$ numbers $\lambda_1,\ldots,\lambda_{n-1}$. i.e.

$$\sigma_k \stackrel{def}{=} \prod_{1\le i_1 < i_2 < \cdots < i_k \le n-1} \lambda_{i_1}\lambda_{i_2}\cdots\lambda_{i_k}$$


$$S_k = \begin{cases} 1, & k = 0\\ \frac{\sigma_k}{\binom{n-1}{k}}, & 0 < k < n \end{cases} $$ be the corresponding elementary symmetry means. It is easy to check

$$\begin{cases} S_{n-1} &= (-1)^{n-1} y_0'\\ S_{n-2} &= \frac{(-1)^{n-2}}{2(n-1)} y_0''\\ S_{n-3} &= \frac{(-1)^{n-3}}{3(n-1)(n-2)} y_0''' \end{cases}\tag{*3} $$ By assumption, all the $\lambda_i$ are real and distinct. By Newton's inequalities, we have $$S_{n-2}^2 \ge S_{n-1}S_{n-3}\tag{*4}$$ Substitute $(*2)$ and $(*3)$ into $(*4)$, this leads to

$$\left(\frac{-\frac{q_0}{p_0}y_0'}{2(n-1)}\right)^2 \ge \frac{-\frac{1}{p_0}\left( (q_0 + p_0')\left(-\frac{q_0}{p_0}\right) + (r_0 + q_0')\right) (y_0')^2}{3(n-1)(n-2)} $$ Notice $y_0', p_0 \ne 0$, we can cancel them in above expression and get

$$\begin{align} & 3(n-2) q_0^2 \ge 4(n-1) \left( (q_0 + p_0') q_0 - (r_0 + q_0') p_0\right)\\ \iff & 4(n-1)\left((r_0 + q_0')p_0 - p_0' q_0\right) - (n+2) q_0^2 \ge 0 \end{align} $$

  • $\begingroup$ Thanks ! I couldn't expect a better answer. $\endgroup$ – Ermite May 9 '15 at 11:14

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