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Math Help - 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 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\}
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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 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.
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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 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) .
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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 |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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