Mixing
Is there a relation of roots for scaled coefficients of real monic polynomial?
Let $pi$ be the bijection between coefficients of the real monic polynomials to the real monic polynomials. Let $ain mathbb R^n$ be fixed vector. Then
begin{align*}
pi(a) = t^n + a_{n-1} t^{n-1} + dots + a_0. \
end{align*}
Let $r > 0$. Is it possible to bound the roots of $pi(ra)$ using the roots $pi(a)$ in the sense: if the roots of $pi(a)$ are in the unit disk, then the roots of $pi(ra)$ would be contained in the unit disk for some suitable choice of $r > 0$? I tried to use Vieta's formula, but it gets quite complicate.
linear-algebra polynomials
asked Nov 20 '18 at 8:41
user1101010
7551630
add a comment |
Let $pi$ be the bijection between coefficients of the real monic polynomials to the real monic polynomials. Let $ain mathbb R^n$ be fixed vector. Then
begin{align*}
pi(a) = t^n + a_{n-1} t^{n-1} + dots + a_0. \
end{align*}
Let $r > 0$. Is it possible to bound the roots of $pi(ra)$ using the roots $pi(a)$ in the sense: if the roots of $pi(a)$ are in the unit disk, then the roots of $pi(ra)$ would be contained in the unit disk for some suitable choice of $r > 0$? I tried to use Vieta's formula, but it gets quite complicate.
linear-algebra polynomials
asked Nov 20 '18 at 8:41
user1101010
7551630
No simple relation that I can see, not even for $n=2$.
– lhf
Nov 20 '18 at 10:34
@lhf: Thanks. Is it possible to bound the roots $pi(ra)$ using the roots $pi(a)$ in the sense: if the roots of $pi(a)$ are in the unit disk, then the roots of $pi(ra)$ would also be in the unit disk for suitable choice of $r > 0$?
– user1101010
Nov 20 '18 at 19:48
1
Add this requirement to your question.
– lhf
Nov 21 '18 at 10:42
With outside instead of inside the unit disk it is easier.
– reuns
Nov 21 '18 at 18:25
add a comment |
Let $pi$ be the bijection between coefficients of the real monic polynomials to the real monic polynomials. Let $ain mathbb R^n$ be fixed vector. Then
begin{align*}
pi(a) = t^n + a_{n-1} t^{n-1} + dots + a_0. \
end{align*}
Let $r > 0$. Is it possible to bound the roots of $pi(ra)$ using the roots $pi(a)$ in the sense: if the roots of $pi(a)$ are in the unit disk, then the roots of $pi(ra)$ would be contained in the unit disk for some suitable choice of $r > 0$? I tried to use Vieta's formula, but it gets quite complicate.
linear-algebra polynomials
asked Nov 20 '18 at 8:41
user1101010
7551630
Let $pi$ be the bijection between coefficients of the real monic polynomials to the real monic polynomials. Let $ain mathbb R^n$ be fixed vector. Then
begin{align*}
pi(a) = t^n + a_{n-1} t^{n-1} + dots + a_0. \
end{align*}
Let $r > 0$. Is it possible to bound the roots of $pi(ra)$ using the roots $pi(a)$ in the sense: if the roots of $pi(a)$ are in the unit disk, then the roots of $pi(ra)$ would be contained in the unit disk for some suitable choice of $r > 0$? I tried to use Vieta's formula, but it gets quite complicate.
linear-algebra polynomials
linear-algebra polynomials
asked Nov 20 '18 at 8:41
user1101010
7551630
asked Nov 20 '18 at 8:41
user1101010
7551630
edited Nov 21 '18 at 18:00
asked Nov 20 '18 at 8:41
user1101010
7551630
asked Nov 20 '18 at 8:41
user1101010
7551630
asked Nov 20 '18 at 8:41
user1101010
7551630
7551630
No simple relation that I can see, not even for $n=2$.
– lhf
Nov 20 '18 at 10:34
@lhf: Thanks. Is it possible to bound the roots $pi(ra)$ using the roots $pi(a)$ in the sense: if the roots of $pi(a)$ are in the unit disk, then the roots of $pi(ra)$ would also be in the unit disk for suitable choice of $r > 0$?
– user1101010
Nov 20 '18 at 19:48
1
Add this requirement to your question.
– lhf
Nov 21 '18 at 10:42
With outside instead of inside the unit disk it is easier.
– reuns
Nov 21 '18 at 18:25
add a comment |
No simple relation that I can see, not even for $n=2$.
– lhf
Nov 20 '18 at 10:34
@lhf: Thanks. Is it possible to bound the roots $pi(ra)$ using the roots $pi(a)$ in the sense: if the roots of $pi(a)$ are in the unit disk, then the roots of $pi(ra)$ would also be in the unit disk for suitable choice of $r > 0$?
– user1101010
Nov 20 '18 at 19:48
1
Add this requirement to your question.
– lhf
Nov 21 '18 at 10:42
With outside instead of inside the unit disk it is easier.
– reuns
Nov 21 '18 at 18:25
No simple relation that I can see, not even for $n=2$.
– lhf
Nov 20 '18 at 10:34
No simple relation that I can see, not even for $n=2$.
– lhf
Nov 20 '18 at 10:34
@lhf: Thanks. Is it possible to bound the roots $pi(ra)$ using the roots $pi(a)$ in the sense: if the roots of $pi(a)$ are in the unit disk, then the roots of $pi(ra)$ would also be in the unit disk for suitable choice of $r > 0$?
– user1101010
Nov 20 '18 at 19:48
@lhf: Thanks. Is it possible to bound the roots $pi(ra)$ using the roots $pi(a)$ in the sense: if the roots of $pi(a)$ are in the unit disk, then the roots of $pi(ra)$ would also be in the unit disk for suitable choice of $r > 0$?
– user1101010
Nov 20 '18 at 19:48
1
1
Add this requirement to your question.
– lhf
Nov 21 '18 at 10:42
Add this requirement to your question.
– lhf
Nov 21 '18 at 10:42
With outside instead of inside the unit disk it is easier.
– reuns
Nov 21 '18 at 18:25
With outside instead of inside the unit disk it is easier.
– reuns
Nov 21 '18 at 18:25
add a comment |
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No simple relation that I can see, not even for $n=2$.
– lhf
Nov 20 '18 at 10:34
@lhf: Thanks. Is it possible to bound the roots $pi(ra)$ using the roots $pi(a)$ in the sense: if the roots of $pi(a)$ are in the unit disk, then the roots of $pi(ra)$ would also be in the unit disk for suitable choice of $r > 0$?
– user1101010
Nov 20 '18 at 19:48
1
Add this requirement to your question.
– lhf
Nov 21 '18 at 10:42
With outside instead of inside the unit disk it is easier.
– reuns
Nov 21 '18 at 18:25