lower bound for absolute values of zeroes of polynomial with non-zero constant term

**ymar** · January 3rd 2012, 01:51 AM

I know there are upper bounds for the absolute value of polynomial zeroes. I have polynomials with non-zero constant terms. How can I find lower bounds for the absolute values of their zeroes? I don't think I need very good bounds, simple ones should do.

**FernandoRevilla** · January 3rd 2012, 07:06 AM

If $f(z)=a_nz^n+\ldots+a_1z+a_0\in\mathbb{C}[z]$ with $a_n\neq 0$ and $c\in \mathbb{C}$ is a root of $f(z)$ then, easily proved $|c|\leq M$ where

$M=\max \left\{\left(n\left |\frac{a_{n-i}}{a_n}\right |\right)^{\frac{1}{i}}:i=1,\ldots,n\right\}$

**ymar** · January 3rd 2012, 12:04 PM

Thank you. I didn't know this bound.

I was asking about something slightly different though. Let $f$ and $c$ be defined as above, but with $a_0\neq 0.$ Now $c\neq 0.$ I'd like to find $M\in \mathbb{R}_+$ such that $c\geq M.$

**FernandoRevilla** · January 3rd 2012, 12:57 PM

Originally Posted by ymar

I was asking about something slightly different though. Let $f$ and $c$ be defined as above, but with $a_0\neq 0.$ Now $c\neq 0.$ I'd like to find $M\in \mathbb{R}_+$ such that $c\geq M.$

Use the substitution $w=1/z$ and you'll have a bound of the form $|c|\geq 1/M_1$ where $M_1$ now corresponds to $z^nf(1/z)$ .

**FernandoRevilla** · January 3rd 2012, 10:45 PM

I provide a proof of the result in answer #2. Suppose $|z|>M$ , then

$|z|>\left(n\left|\dfrac{a_{i-1}}{a_n}\right|\right)^{\frac{1}{i}}\;(\forall i=1,\ldots,n)\Rightarrow |z|^i>n\dfrac{|a_{i-1}|}{|a_n|}\;\;(\forall i=1,\ldots,n)\Rightarrow$

$|a_{n-i}|<\dfrac{|a_n||z|^i}{n} \;\;(\forall i=1,\ldots,n)\Rightarrow|a_{n-1}z^{n-1}+\ldots+a_1z+a_0|\leq$

$|a_{n-1}||z|^{n-1}+\ldots+|a_1||z|+|a_0|<\dfrac{|a_n||z|^n}{n}+\;. ..\;+\dfrac{|a_n||z|^n}{n}=|a_nz^n|$

That is, if $|z|>M$ then $|a_{n-1}z^{n-1}+\ldots+a_1z+a_0|<|-a_nz^n|$ , so $z$ is not a root of $f(z)$ .

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