IMC 2006 · Day 1 · P6

Let $A(x) = a_0 + a_1 x + \dots + a_n x^n$. We prove that sequence $a_0, \dots, a_n$ satisfies the required property if and only if all zeros of polynomial $A(x)$ are real.

(a) Assume that all roots of $A(x)$ are real. Let us use the following notations. Let $I$ be the identity operator on $\mathbb{R} \to \mathbb{R}$ functions and $D$ be differentiation operator. For an arbitrary polynomial $P(x) = p_0 + p_1 x + \dots + p_n x^n$, write $P(D) = p_0 I + p_1 D + p_2 D^2 + \dots + p_n D^n$. Then the statement can written as $(A(D) f)(\xi) = 0$.

First prove the statement for $n = 1$. Consider the function $$g(x) = e^{\frac{a_0}{a_1} x} f(x).$$ Since $g(x_0) = g(x_1) = 0$, by Rolle's theorem there exists a $\xi \in (x_0, x_1)$ for which $$g'(\xi) = e^{\frac{a_0}{a_1} \xi} \frac{a_0}{a_1} f(\xi) + e^{\frac{a_0}{a_1} \xi} f'(\xi) = \frac{e^{\frac{a_0}{a_1} \xi}}{a_1} (a_0 f(\xi) + a_1 f'(\xi)) = 0.$$ Now assume that $n > 1$ and the statement holds for $n-1$. Let $A(x) = (x - c) B(x)$ where $c$ is a real root of polynomial $A$. By the $n = 1$ case, there exist $y_0 \in (x_0, x_1)$, $y_1 \in (x_1, x_2)$, …, $y_{n-1} \in (x_{n-1}, x_n)$ such that $f'(y_j) - c f(y_j) = 0$ for all $j = 0, 1, \dots, n-1$. Now apply the induction hypothesis for polynomial $B(x)$, function $g = f' - cf$ and points $y_0, \dots, y_{n-1}$. The hypothesis says that there exists a $\xi \in (y_0, y_{n-1}) \subset (x_0, x_n)$ such that $$(B(D) g)(\xi) = (B(D)(D - cI) f)(\xi) = (A(D) f)(\xi) = 0.$$

(b) Assume that $u + vi$ is a complex root of polynomial $A(x)$ such that $v \ne 0$. Consider the linear differential equation $a_n g^{(n)} + \dots + a_1 g' + g = 0$.

A solution of this equation is $g_1(x) = e^{ux} \sin vx$ which has infinitely many zeros.

Let $k$ be the smallest index for which $a_k \ne 0$. Choose a small $\varepsilon > 0$ and set $f(x) = g_1(x) + \varepsilon x^k$. If $\varepsilon$ is sufficiently small then $g$ has the required number of roots but $a_0 f + a_1 f' + \dots + a_n f^{(n)} = a_k \varepsilon \ne 0$

everywhere.