3
$\begingroup$

Problem: Let $a_0, a_1, \ldots, a_{n-1}$ be real numbers where $n \geq 1$ and let $f(x) = x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \ldots + a_0$ be such that $|f(0)| = f(1)$ and each root of $f(x) = 0$ is real and lies between $0$ to $1$. Prove that the product of the roots of this equation does not exceed $\frac{1}{2^n}$

This problem was brought to my attention through a friend. I could only obtain weak bounds by using AM-GM along with Vieta's lemma, which is nowhere close to the bound required. I thought of some relation with sum of binomial coefficients of $(1+x)^n$, where $x = 1$, but that also got me nowhere.

Any help with this problem would be appreciated.

$\endgroup$

1 Answer 1

8
$\begingroup$

Let $f(x) = (x-\lambda_1)(x-\lambda_2)\dots (x-\lambda_n)$, where $0 \leq \lambda_1 \leq \dots \leq \lambda_n \leq 1$ are the roots of $f$.

The product of the roots is $\pi_n := \lambda_1 \dots \lambda_n = |f(0)|$, and we also have $|f(0)| = f(1) = (1-\lambda_1) \dots (1-\lambda_n)$.

Then we have: $$\pi_n^2 = |f(0)|f(1) = \lambda_1(1-\lambda_1)\dots \lambda_n(1-\lambda_n) \leq \frac{1}{4^n}$$

Where the upper bound comes from the fact that $x(1-x) \leq 1/4$ for $x \in [0,1]$.

Therefore we deduce that $\pi_n \leq 1/2^n$.

$\endgroup$
1
  • $\begingroup$ Nice observation, I didn't notice that! $\endgroup$ Aug 7, 2023 at 11:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .