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Question Number 216800 by depressiveshrek last updated on 21/Feb/25

	Answered by MrGaster last updated on 21/Feb/25

	$Prove : f (x) = a {(x - r_{1})}^{m_{1}} {(x - r_{2})}^{m_{2}} \dots {(x - r_{k})}^{m_{k}}$ $where a > 0, r_{1}, r_{2}, \dots, r_{k} are the distinct$ $real roots of f (x), and m_{1}, m_{2}, \dots m_{k} are$ $their respective multiplicities (1)$ $The derivatives of f (x) are :$ $f^{(j)} (x) = a \underset{i = 1}{\sum^{k}} \frac{m_{i}!}{(m_{i} - j)!} {(x - r_{i})}^{m_{i} - j} \prod_{l \neq i} {(x - r_{i})}^{m_{l}}$ $for j = 1, 2, \dots, n .$ $g (x) :$ $g (x) = f (x) + f^{'} (x) \dots + f^{(n)} (x)$ $Substituting the expressionsfor thes$ $derivative have :$ $g (x) = a \underset{j = 0}{\sum^{n}} \underset{i = 1}{\sum^{k}} \frac{m_{i}!}{(m_{i} - j)} {(x - r_{i})}^{m_{i} - j} \prod_{l \neq j} {(x - r_{l})}^{m_{l}} (2)$ $since f (x) \geq 0 for all x, each term in the sum f (x) + f^{'} (x) + \dots + f^{(n)} is nonnegative . Specificallyr$ $fo each root r_{i}, the terms involving {(x - r_{i})}^{m - j} for j = 0, 1, \dots m_{i} are nonnegative because m_{i} - j \geq 0$ $Therefore the sum of theser$ $tems is also nonnegative . c$ $Hene, g (x) is a sum of nonnegatives$ $term which implies :$ $\begin{matrix} g (x) \geq 0 \end{matrix}$ $[Q . E . D]$
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