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

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    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.
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    MHF Contributor FernandoRevilla's Avatar
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    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

    $\displaystyle M=\max \left\{\left(n\left |\frac{a_{n-i}}{a_n}\right |\right)^{\frac{1}{i}}:i=1,\ldots,n\right\}$
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    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 $\displaystyle c\geq M.$
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    MHF Contributor FernandoRevilla's Avatar
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    Re: lower bound for absolute values of zeroes of polynomial with non-zero constant te

    Quote Originally Posted by ymar View Post
    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 $\displaystyle z^nf(1/z)$ .
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    MHF Contributor FernandoRevilla's Avatar
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    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 $\displaystyle f(z)$ .
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