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How to prove that all the roots of the polynomial $f(x)=a_o+a_1 x+\cdots+ x^n$ with real coefficients belong to the interval $ [-M, M] $, with $\displaystyle{M=1+\sum_{k=0}^{n}|a_k|}$

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Hi Jonas. I honestly thought the exercise but not to use: ( –  Roiner Segura Cubero Dec 7 '12 at 3:56
Induction tried but did not succeed –  Roiner Segura Cubero Dec 7 '12 at 4:00
Your accept rate is too low. I'd do something about it. –  DonAntonio Dec 7 '12 at 4:11
my accept rate? –  Roiner Segura Cubero Dec 7 '12 at 4:13
As written the claim is false, since the roots of $\,x^2+1\in\Bbb R[x]\,$ do not belong to the interval $\,[-3,3]\,$ ...Are there any other conditions on the polynomial? –  DonAntonio Dec 7 '12 at 4:14
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2 Answers 2

But let me give you a slightly different (and in a sense, stronger) formulation that is true.

Claim: if $P(x) = x^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0$, then a real root $r$ of $P$ satisfies $|r| \leq 1 + \max|a_i|$.

Suppose WLOG that $|r|>1$. Then $P(r) = 0$ means that $r^n = -a_0 - a_1r - \ldots - a_{n-1}r^{n-1}$. Let's call $M = \max|a_i|$. Then since $|r^n| \leq |a_0| + |a_1||r| + \ldots + |a_{n-1}||r^{n-1}|$, we have that $|r|^n \leq M(1 + |r| + \ldots + |r|^{n-1})$. This is a finite geometric series, and writing this out gives that $|x|-1 \leq M(1 - \frac{1}{|r|^n}) \leq M$, as was to be proven. $\diamondsuit$

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If $|x|\ge1$ and $f(x)=0$ then $$|x^n|=|-a_0-a_1x-\cdots-a_{n-1}x^{n-1}|\le|x|^{n-1}\sum|a_i|$$

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