Can the max of a polynomial be near a zero?
$begingroup$
The setup of this question is very similar to a recent question I asked.
Let $p(z)$ be a polynomial of degree $n$ with complex zeros ${z_i}_{i=1}^n subsetmathbb{C}$. Let $Zsubset mathbb{C}$ be the set near the zeros,
$$
Z = bigcup_{i=1}^n D_{(100n)^{-1}}(z_i),
$$
where $D_r(x)subset mathbb{C}$ is the disk of radius $r$ centered at $x$. Let $gamma$ be the line segment $[-1,1]$ on the real axis in the complex plane.
I am interested in whether the inequality
$$
sup_{gammacap Z} |p(x)| leq sup_{xin gammasetminus Z} |p(x)|
$$
holds. In other words, is it possible that the maximum value of $p$ on the line segment $gamma$ takes place in $Z$?
It seems that the maximum principle might be useful here, but I am only interested in what happens to the polynomial on a line segment (which does not contain any disks in the complex plane). I am not sure how to proceed otherwise.
real-analysis complex-analysis polynomials
asked Dec 24 '18 at 15:34
felipehfelipeh
1,616619
$endgroup$
add a comment |
$begingroup$
The setup of this question is very similar to a recent question I asked.
Let $p(z)$ be a polynomial of degree $n$ with complex zeros ${z_i}_{i=1}^n subsetmathbb{C}$. Let $Zsubset mathbb{C}$ be the set near the zeros,
$$
Z = bigcup_{i=1}^n D_{(100n)^{-1}}(z_i),
$$
where $D_r(x)subset mathbb{C}$ is the disk of radius $r$ centered at $x$. Let $gamma$ be the line segment $[-1,1]$ on the real axis in the complex plane.
I am interested in whether the inequality
$$
sup_{gammacap Z} |p(x)| leq sup_{xin gammasetminus Z} |p(x)|
$$
holds. In other words, is it possible that the maximum value of $p$ on the line segment $gamma$ takes place in $Z$?
It seems that the maximum principle might be useful here, but I am only interested in what happens to the polynomial on a line segment (which does not contain any disks in the complex plane). I am not sure how to proceed otherwise.
real-analysis complex-analysis polynomials
asked Dec 24 '18 at 15:34
felipehfelipeh
1,616619
$endgroup$
add a comment |
$begingroup$
The setup of this question is very similar to a recent question I asked.
Let $p(z)$ be a polynomial of degree $n$ with complex zeros ${z_i}_{i=1}^n subsetmathbb{C}$. Let $Zsubset mathbb{C}$ be the set near the zeros,
$$
Z = bigcup_{i=1}^n D_{(100n)^{-1}}(z_i),
$$
where $D_r(x)subset mathbb{C}$ is the disk of radius $r$ centered at $x$. Let $gamma$ be the line segment $[-1,1]$ on the real axis in the complex plane.
I am interested in whether the inequality
$$
sup_{gammacap Z} |p(x)| leq sup_{xin gammasetminus Z} |p(x)|
$$
holds. In other words, is it possible that the maximum value of $p$ on the line segment $gamma$ takes place in $Z$?
It seems that the maximum principle might be useful here, but I am only interested in what happens to the polynomial on a line segment (which does not contain any disks in the complex plane). I am not sure how to proceed otherwise.
real-analysis complex-analysis polynomials
asked Dec 24 '18 at 15:34
felipehfelipeh
1,616619
$endgroup$
The setup of this question is very similar to a recent question I asked.
Let $p(z)$ be a polynomial of degree $n$ with complex zeros ${z_i}_{i=1}^n subsetmathbb{C}$. Let $Zsubset mathbb{C}$ be the set near the zeros,
$$
Z = bigcup_{i=1}^n D_{(100n)^{-1}}(z_i),
$$
where $D_r(x)subset mathbb{C}$ is the disk of radius $r$ centered at $x$. Let $gamma$ be the line segment $[-1,1]$ on the real axis in the complex plane.
I am interested in whether the inequality
$$
sup_{gammacap Z} |p(x)| leq sup_{xin gammasetminus Z} |p(x)|
$$
holds. In other words, is it possible that the maximum value of $p$ on the line segment $gamma$ takes place in $Z$?
It seems that the maximum principle might be useful here, but I am only interested in what happens to the polynomial on a line segment (which does not contain any disks in the complex plane). I am not sure how to proceed otherwise.
real-analysis complex-analysis polynomials
real-analysis complex-analysis polynomials
asked Dec 24 '18 at 15:34
felipehfelipeh
1,616619
asked Dec 24 '18 at 15:34
felipehfelipeh
1,616619
asked Dec 24 '18 at 15:34
felipehfelipeh
1,616619
asked Dec 24 '18 at 15:34
felipehfelipeh
1,616619
asked Dec 24 '18 at 15:34
felipehfelipeh
1,616619
1,616619
add a comment |
add a comment |
1 Answer
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$begingroup$
The answer is no! Consider the Chebyshev polynomial $T_n(x)$ with roots
$$
x_k = cosleft(frac{2k-1}{2n}piright),
$$
and which satisfies $|T_n(x)| leq T_n(1)$. The $n$-th root is a distance $Theta(n^{-2})$ away from the maximum value.
answered Jan 2 at 18:11
felipehfelipeh
1,616619
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer is no! Consider the Chebyshev polynomial $T_n(x)$ with roots
$$
x_k = cosleft(frac{2k-1}{2n}piright),
$$
and which satisfies $|T_n(x)| leq T_n(1)$. The $n$-th root is a distance $Theta(n^{-2})$ away from the maximum value.
answered Jan 2 at 18:11
felipehfelipeh
1,616619
$endgroup$
add a comment |
$begingroup$
The answer is no! Consider the Chebyshev polynomial $T_n(x)$ with roots
$$
x_k = cosleft(frac{2k-1}{2n}piright),
$$
and which satisfies $|T_n(x)| leq T_n(1)$. The $n$-th root is a distance $Theta(n^{-2})$ away from the maximum value.
answered Jan 2 at 18:11
felipehfelipeh
1,616619
$endgroup$
add a comment |
$begingroup$
The answer is no! Consider the Chebyshev polynomial $T_n(x)$ with roots
$$
x_k = cosleft(frac{2k-1}{2n}piright),
$$
and which satisfies $|T_n(x)| leq T_n(1)$. The $n$-th root is a distance $Theta(n^{-2})$ away from the maximum value.
answered Jan 2 at 18:11
felipehfelipeh
1,616619
$endgroup$
The answer is no! Consider the Chebyshev polynomial $T_n(x)$ with roots
$$
x_k = cosleft(frac{2k-1}{2n}piright),
$$
and which satisfies $|T_n(x)| leq T_n(1)$. The $n$-th root is a distance $Theta(n^{-2})$ away from the maximum value.
answered Jan 2 at 18:11
felipehfelipeh
1,616619
answered Jan 2 at 18:11
felipehfelipeh
1,616619
answered Jan 2 at 18:11
felipehfelipeh
1,616619
answered Jan 2 at 18:11
felipehfelipeh
1,616619
1,616619
add a comment |
add a comment |
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