Mixing
Writing the roots of a polynomial with varying coefficients as continuous functions?
$begingroup$
Consider the monic polynomial
$$p_{zeta}(z) = z^n + a_{n-1}(zeta)z^{n-1} + dots + a_0(zeta), $$
where the $a_{i}$'s are continuous functions defined over $mathbb{C}$. As is well known, the roots of this polynomial depend 'continuously' on the $a_{i}$'s in some sense. However, I was wondering if a stronger statement holds: is it possible to pick continuous functions $f_1, dots, f_n$, at least locally, such that the roots of $p_{zeta}(z)$ are exactly $f_1(zeta), dots, f_n(zeta)$, counting multiplicities?
Edit: Note this is easily answered in the affirmative around a $zeta_0$ such that $p_{zeta_0}(z)$ has $n$ distinct roots. The real difficulty is in dealing with the case where $p_{zeta}(z)$ has repeated roots.
complex-analysis analysis algebraic-geometry polynomials roots
asked Sep 21 '14 at 20:14
user18063user18063
757314
$endgroup$
add a comment |
$begingroup$
Consider the monic polynomial
$$p_{zeta}(z) = z^n + a_{n-1}(zeta)z^{n-1} + dots + a_0(zeta), $$
where the $a_{i}$'s are continuous functions defined over $mathbb{C}$. As is well known, the roots of this polynomial depend 'continuously' on the $a_{i}$'s in some sense. However, I was wondering if a stronger statement holds: is it possible to pick continuous functions $f_1, dots, f_n$, at least locally, such that the roots of $p_{zeta}(z)$ are exactly $f_1(zeta), dots, f_n(zeta)$, counting multiplicities?
Edit: Note this is easily answered in the affirmative around a $zeta_0$ such that $p_{zeta_0}(z)$ has $n$ distinct roots. The real difficulty is in dealing with the case where $p_{zeta}(z)$ has repeated roots.
complex-analysis analysis algebraic-geometry polynomials roots
asked Sep 21 '14 at 20:14
user18063user18063
757314
$endgroup$
$begingroup$
This could be related math.stackexchange.com/questions/63196/…
$endgroup$
– PhoemueX
Sep 21 '14 at 21:03
$begingroup$
Isn't this just a combination of the well-known "roots depend continously on the coefficients" and the fact that continuous functions are closed under composition?
$endgroup$
– Martin Brandenburg
Sep 21 '14 at 21:22
add a comment |
$begingroup$
Consider the monic polynomial
$$p_{zeta}(z) = z^n + a_{n-1}(zeta)z^{n-1} + dots + a_0(zeta), $$
where the $a_{i}$'s are continuous functions defined over $mathbb{C}$. As is well known, the roots of this polynomial depend 'continuously' on the $a_{i}$'s in some sense. However, I was wondering if a stronger statement holds: is it possible to pick continuous functions $f_1, dots, f_n$, at least locally, such that the roots of $p_{zeta}(z)$ are exactly $f_1(zeta), dots, f_n(zeta)$, counting multiplicities?
Edit: Note this is easily answered in the affirmative around a $zeta_0$ such that $p_{zeta_0}(z)$ has $n$ distinct roots. The real difficulty is in dealing with the case where $p_{zeta}(z)$ has repeated roots.
complex-analysis analysis algebraic-geometry polynomials roots
asked Sep 21 '14 at 20:14
user18063user18063
757314
$endgroup$
Consider the monic polynomial
$$p_{zeta}(z) = z^n + a_{n-1}(zeta)z^{n-1} + dots + a_0(zeta), $$
where the $a_{i}$'s are continuous functions defined over $mathbb{C}$. As is well known, the roots of this polynomial depend 'continuously' on the $a_{i}$'s in some sense. However, I was wondering if a stronger statement holds: is it possible to pick continuous functions $f_1, dots, f_n$, at least locally, such that the roots of $p_{zeta}(z)$ are exactly $f_1(zeta), dots, f_n(zeta)$, counting multiplicities?
Edit: Note this is easily answered in the affirmative around a $zeta_0$ such that $p_{zeta_0}(z)$ has $n$ distinct roots. The real difficulty is in dealing with the case where $p_{zeta}(z)$ has repeated roots.
complex-analysis analysis algebraic-geometry polynomials roots
complex-analysis analysis algebraic-geometry polynomials roots
asked Sep 21 '14 at 20:14
user18063user18063
757314
asked Sep 21 '14 at 20:14
user18063user18063
757314
edited Sep 21 '14 at 22:29
user18063
asked Sep 21 '14 at 20:14
user18063user18063
757314
asked Sep 21 '14 at 20:14
user18063user18063
757314
asked Sep 21 '14 at 20:14
user18063user18063
757314
757314
$begingroup$
This could be related math.stackexchange.com/questions/63196/…
$endgroup$
– PhoemueX
Sep 21 '14 at 21:03
$begingroup$
Isn't this just a combination of the well-known "roots depend continously on the coefficients" and the fact that continuous functions are closed under composition?
$endgroup$
– Martin Brandenburg
Sep 21 '14 at 21:22
add a comment |
$begingroup$
This could be related math.stackexchange.com/questions/63196/…
$endgroup$
– PhoemueX
Sep 21 '14 at 21:03
$begingroup$
Isn't this just a combination of the well-known "roots depend continously on the coefficients" and the fact that continuous functions are closed under composition?
$endgroup$
– Martin Brandenburg
Sep 21 '14 at 21:22
$begingroup$
This could be related math.stackexchange.com/questions/63196/…
$endgroup$
– PhoemueX
Sep 21 '14 at 21:03
$begingroup$
This could be related math.stackexchange.com/questions/63196/…
$endgroup$
– PhoemueX
Sep 21 '14 at 21:03
$begingroup$
Isn't this just a combination of the well-known "roots depend continously on the coefficients" and the fact that continuous functions are closed under composition?
$endgroup$
– Martin Brandenburg
Sep 21 '14 at 21:22
$begingroup$
Isn't this just a combination of the well-known "roots depend continously on the coefficients" and the fact that continuous functions are closed under composition?
$endgroup$
– Martin Brandenburg
Sep 21 '14 at 21:22
add a comment |
1 Answer
1
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$begingroup$
No, this is not possible, and the example is $z^2-zeta=0$. Even if you restrict $zeta$ on the unit circle, there are evidently no continuous functions $z_1(zeta)$ and $z_2(zeta)$ defined for all $zeta$ on the unit circle,
and giving the two roots of this equation. Even ONE continuous function $z(zeta)$
on the unit circle, that satisfies $z^2(zeta)=zeta$ evidently does not exist.
answered Mar 29 '15 at 17:29
Alexandre EremenkoAlexandre Eremenko
2,455921
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
No, this is not possible, and the example is $z^2-zeta=0$. Even if you restrict $zeta$ on the unit circle, there are evidently no continuous functions $z_1(zeta)$ and $z_2(zeta)$ defined for all $zeta$ on the unit circle,
and giving the two roots of this equation. Even ONE continuous function $z(zeta)$
on the unit circle, that satisfies $z^2(zeta)=zeta$ evidently does not exist.
answered Mar 29 '15 at 17:29
Alexandre EremenkoAlexandre Eremenko
2,455921
$endgroup$
add a comment |
$begingroup$
No, this is not possible, and the example is $z^2-zeta=0$. Even if you restrict $zeta$ on the unit circle, there are evidently no continuous functions $z_1(zeta)$ and $z_2(zeta)$ defined for all $zeta$ on the unit circle,
and giving the two roots of this equation. Even ONE continuous function $z(zeta)$
on the unit circle, that satisfies $z^2(zeta)=zeta$ evidently does not exist.
answered Mar 29 '15 at 17:29
Alexandre EremenkoAlexandre Eremenko
2,455921
$endgroup$
add a comment |
$begingroup$
No, this is not possible, and the example is $z^2-zeta=0$. Even if you restrict $zeta$ on the unit circle, there are evidently no continuous functions $z_1(zeta)$ and $z_2(zeta)$ defined for all $zeta$ on the unit circle,
and giving the two roots of this equation. Even ONE continuous function $z(zeta)$
on the unit circle, that satisfies $z^2(zeta)=zeta$ evidently does not exist.
answered Mar 29 '15 at 17:29
Alexandre EremenkoAlexandre Eremenko
2,455921
$endgroup$
No, this is not possible, and the example is $z^2-zeta=0$. Even if you restrict $zeta$ on the unit circle, there are evidently no continuous functions $z_1(zeta)$ and $z_2(zeta)$ defined for all $zeta$ on the unit circle,
and giving the two roots of this equation. Even ONE continuous function $z(zeta)$
on the unit circle, that satisfies $z^2(zeta)=zeta$ evidently does not exist.
answered Mar 29 '15 at 17:29
Alexandre EremenkoAlexandre Eremenko
2,455921
edited Apr 1 '15 at 2:40
answered Mar 29 '15 at 17:29
Alexandre EremenkoAlexandre Eremenko
2,455921
answered Mar 29 '15 at 17:29
Alexandre EremenkoAlexandre Eremenko
2,455921
answered Mar 29 '15 at 17:29
Alexandre EremenkoAlexandre Eremenko
2,455921
2,455921
add a comment |
add a comment |
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$begingroup$
This could be related math.stackexchange.com/questions/63196/…
$endgroup$
– PhoemueX
Sep 21 '14 at 21:03
$begingroup$
Isn't this just a combination of the well-known "roots depend continously on the coefficients" and the fact that continuous functions are closed under composition?
$endgroup$
– Martin Brandenburg
Sep 21 '14 at 21:22