A Problem from “Problem Solving through Problems”

on May 2, 2018

One of my favourite mathematical books is "Problem Solving Through Problems" by Loren C. Larson. It contains a wealth of techniques and problems, many of which appeared in previous Putman competitions (an annual mathematical competition for undergraduate students). It also has many exercises. The following is one of my favourite problems from that book, with its solution that took me a long time to get. The interested reader should try to solve this before looking at the solution.

Problem: Let $P(x)$ be a polynomial of degree $n$ such that $P(t) \geq 0$ for all $t \in \mathbb{R}$ . Show that

$P(t)+P'(t)+\ldots+P^{(n)}(t) \geq 0$

for all $t \in \mathbb{R}$ . (Here $P^{(n)}(t)$ refers to the $n$ -th derivative of $P(x)$ evaluated at $t$ .)

Solution: Note that $P^{(n+1)}(x)$ is identically 0, since taking a derivative lowers the degree by 1. Let

$Q(x) = P(x)+P'(x)+\ldots+P^{(n)}(x).$

Then

$Q'(x) = P'(x)+\ldots+P^{(n)}(x),$

by the above observation. Note that $Q(x)-Q'(x)=P(x)$ .

Now $P(x)$ is of even degree since $P(t) \geq 0$ for all $t$ . Thus by definition, $Q(x)$ is also of even degree. This implies $Q(x)$ has a minimum. Let us suppose it attains its minimum at $t_0$ . Then $Q'(t_0)=0$ and so $Q(t) \geq Q(t_0) = P(t_0) \geq 0$ , for all $t$ , which shows the required result.

Categories:

Mathematics

Tags:

No Tag

Comments are closed