Real polynomial

Moderator: elim

Real polynomial

Let $P(x) \in \mathbb{R}[x]$ having degree $n$ and only real zeros.
Prove that $(n-1) \left( P'(x) \right)^2 \geq nP(x) P''(x)$ for all $x \in \mathbb{R}$

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elim

Posts: 64
Joined: Mon Feb 08, 2010 8:03 pm

Re: Real polynomial

Both sides are $0$ if $n=0, 1$ so assume $n>1$. Let the zeros of $P(x)$ be $x_1, x_2, .. x_n$. The equality is clearly true when $x=x_i$ is a root, with equality if $P'(x_i)=0$, so that $x_i$ is a multiple zero of $P(x)$. Now, if $x$ is not a root of $P$ then we can apply the well known identities: $\displaystyle{\frac{P'(x)}{P(x)} = \sum_{i=1}^n \frac{1}{x-x_i} }$, $\displaystyle{\frac{P''(x)}{P(x)} = \sum_{1\leq i < j \leq n} \frac{2}{(x-x_i)(x-x_j) } }$. They yield $\displaystyle{(n-1) \left( \frac{P'(x)}{P(x)} \right)^2 - n \frac{P''(x)}{P(x)} }$ $\displaystyle{ = \sum_{i=1}^n \frac{n-1}{(x-x_i)^2} - \sum_{1\leq i <j \leq n}\frac{2}{(x-x_i)(x-x_j)} }$ $\displaystyle{= \sum_{1\leq i < j \leq n} \left( \frac{1}{x-x_i} - \frac{1}{x-x_j} \right)^2 }$ which proves the inequality. The equality we require $x_1=x_2=...=x_n$ and a quick check shows $P(x) = C(x-x_1)^n$ indeed gives equality.
elim

Posts: 64
Joined: Mon Feb 08, 2010 8:03 pm

nice post

good to know, thanks for sharing

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seamsql

Posts: 2
Joined: Fri Sep 02, 2011 8:11 am