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

• Jan 3rd 2012, 01:51 AM
ymar
lower bound for absolute values of zeroes of polynomial with non-zero constant term
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.
• Jan 3rd 2012, 07:06 AM
FernandoRevilla
Re: lower bound for absolute values of zeroes of polynomial with non-zero constant te
If $\displaystyle f(z)=a_nz^n+\ldots+a_1z+a_0\in\mathbb{C}[z]$ with $\displaystyle a_n\neq 0$ and $\displaystyle c\in \mathbb{C}$ is a root of $\displaystyle f(z)$ then, easily proved $\displaystyle |c|\leq M$ where

$\displaystyle M=\max \left\{\left(n\left |\frac{a_{n-i}}{a_n}\right |\right)^{\frac{1}{i}}:i=1,\ldots,n\right\}$
• Jan 3rd 2012, 12:04 PM
ymar
Re: lower bound for absolute values of zeroes of polynomial with non-zero constant te
Thank you. I didn't know this bound.

I was asking about something slightly different though. Let $\displaystyle f$ and $\displaystyle c$ be defined as above, but with $\displaystyle a_0\neq 0.$ Now $\displaystyle c\neq 0.$ I'd like to find $\displaystyle M\in \mathbb{R}_+$ such that $\displaystyle c\geq M.$
• Jan 3rd 2012, 12:57 PM
FernandoRevilla
Re: lower bound for absolute values of zeroes of polynomial with non-zero constant te
Quote:

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

Use the substitution $\displaystyle w=1/z$ and you'll have a bound of the form $\displaystyle |c|\geq 1/M_1$ where $\displaystyle M_1$ now corresponds to $\displaystyle z^nf(1/z)$ .
• Jan 3rd 2012, 10:45 PM
FernandoRevilla
Re: lower bound for absolute values of zeroes of polynomial with non-zero constant te
I provide a proof of the result in answer #2. Suppose $\displaystyle |z|>M$ , then

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

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

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

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