Mixing
If coefficients are polynomial functions over $mathbb R$ of a monic polynomial, can we find $n$ continuous...
$begingroup$
$newcommand{a}{alpha} newcommand{bb}{mathbb} newcommand{b}{beta}$
Let
begin{align*}
p(x, t) = x^n + a_{n-1}(t) x^{n-1} + dots + a_1(t) x + a_0
end{align*}
be a monic polynomial where coefficients ${a_0(t), dots, a_{n-1}(t)}$ are real-valued continuous functions over $t in bb R$. In particular, each $a_j(t)$ is polynomial in $t$ with real coefficients.
My question is: could we be able to find $n$ continuous complex-valued functions ${b_0(t), dots, b_{n-1}(t)}$ over $t in mathbb R$ such that for each $t$, ${b_j(t)}$ constitute the roots of the monic polynomial $x^n + a_{n-1}(t) x^{n-1} + dots + a_1(t) x + a_0$? I think the answer is positive since we are working over domain $mathbb R$. If this is true, are these functions polynomials in $t$ (probably with complex coefficients)?
linear-algebra general-topology polynomials
asked Nov 27 '18 at 17:54
user1101010user1101010
9011830
$endgroup$
|
show 16 more comments
$begingroup$
$newcommand{a}{alpha} newcommand{bb}{mathbb} newcommand{b}{beta}$
Let
begin{align*}
p(x, t) = x^n + a_{n-1}(t) x^{n-1} + dots + a_1(t) x + a_0
end{align*}
be a monic polynomial where coefficients ${a_0(t), dots, a_{n-1}(t)}$ are real-valued continuous functions over $t in bb R$. In particular, each $a_j(t)$ is polynomial in $t$ with real coefficients.
My question is: could we be able to find $n$ continuous complex-valued functions ${b_0(t), dots, b_{n-1}(t)}$ over $t in mathbb R$ such that for each $t$, ${b_j(t)}$ constitute the roots of the monic polynomial $x^n + a_{n-1}(t) x^{n-1} + dots + a_1(t) x + a_0$? I think the answer is positive since we are working over domain $mathbb R$. If this is true, are these functions polynomials in $t$ (probably with complex coefficients)?
linear-algebra general-topology polynomials
asked Nov 27 '18 at 17:54
user1101010user1101010
9011830
$endgroup$
1
$begingroup$
The functions $beta_i$ will not be polynomials in general. Consider for instance $p(x,t)=x^2+t$.
$endgroup$
– Federico
Nov 27 '18 at 18:03
$begingroup$
Moreover, see math.stackexchange.com/questions/940653/…
$endgroup$
– Federico
Nov 27 '18 at 18:05
$begingroup$
@Federico: Thanks. I see the functions could not be polynomial. But do they exist? The link was considering the domain $mathbb C$ whereas here the domain of interest is $mathbb R$.
$endgroup$
– user1101010
Nov 27 '18 at 18:26
$begingroup$
The roots of $p(,cdot,,t)$ are a continuous function $mathbb Rto mathbb C^n/mathrm{perm}$, where the quotient is w.r.t. permutations. The fact that the domain is $mathbb R$ implies that you can lift this to a continuous function $beta:mathbb Rtomathbb R^n$ representing the roots
$endgroup$
– Federico
Nov 27 '18 at 18:27
$begingroup$
Basically, you have troubles only when multiple roots coincide. When they separate again (if they do), you just decide arbitrarily which $beta_i$ tracks which root
$endgroup$
– Federico
Nov 27 '18 at 18:30
|
show 16 more comments
$begingroup$
$newcommand{a}{alpha} newcommand{bb}{mathbb} newcommand{b}{beta}$
Let
begin{align*}
p(x, t) = x^n + a_{n-1}(t) x^{n-1} + dots + a_1(t) x + a_0
end{align*}
be a monic polynomial where coefficients ${a_0(t), dots, a_{n-1}(t)}$ are real-valued continuous functions over $t in bb R$. In particular, each $a_j(t)$ is polynomial in $t$ with real coefficients.
My question is: could we be able to find $n$ continuous complex-valued functions ${b_0(t), dots, b_{n-1}(t)}$ over $t in mathbb R$ such that for each $t$, ${b_j(t)}$ constitute the roots of the monic polynomial $x^n + a_{n-1}(t) x^{n-1} + dots + a_1(t) x + a_0$? I think the answer is positive since we are working over domain $mathbb R$. If this is true, are these functions polynomials in $t$ (probably with complex coefficients)?
linear-algebra general-topology polynomials
asked Nov 27 '18 at 17:54
user1101010user1101010
9011830
$endgroup$
$newcommand{a}{alpha} newcommand{bb}{mathbb} newcommand{b}{beta}$
Let
begin{align*}
p(x, t) = x^n + a_{n-1}(t) x^{n-1} + dots + a_1(t) x + a_0
end{align*}
be a monic polynomial where coefficients ${a_0(t), dots, a_{n-1}(t)}$ are real-valued continuous functions over $t in bb R$. In particular, each $a_j(t)$ is polynomial in $t$ with real coefficients.
My question is: could we be able to find $n$ continuous complex-valued functions ${b_0(t), dots, b_{n-1}(t)}$ over $t in mathbb R$ such that for each $t$, ${b_j(t)}$ constitute the roots of the monic polynomial $x^n + a_{n-1}(t) x^{n-1} + dots + a_1(t) x + a_0$? I think the answer is positive since we are working over domain $mathbb R$. If this is true, are these functions polynomials in $t$ (probably with complex coefficients)?
linear-algebra general-topology polynomials
linear-algebra general-topology polynomials
asked Nov 27 '18 at 17:54
user1101010user1101010
9011830
asked Nov 27 '18 at 17:54
user1101010user1101010
9011830
asked Nov 27 '18 at 17:54
user1101010user1101010
9011830
asked Nov 27 '18 at 17:54
user1101010user1101010
9011830
asked Nov 27 '18 at 17:54
user1101010user1101010
9011830
9011830
1
$begingroup$
The functions $beta_i$ will not be polynomials in general. Consider for instance $p(x,t)=x^2+t$.
$endgroup$
– Federico
Nov 27 '18 at 18:03
$begingroup$
Moreover, see math.stackexchange.com/questions/940653/…
$endgroup$
– Federico
Nov 27 '18 at 18:05
$begingroup$
@Federico: Thanks. I see the functions could not be polynomial. But do they exist? The link was considering the domain $mathbb C$ whereas here the domain of interest is $mathbb R$.
$endgroup$
– user1101010
Nov 27 '18 at 18:26
$begingroup$
The roots of $p(,cdot,,t)$ are a continuous function $mathbb Rto mathbb C^n/mathrm{perm}$, where the quotient is w.r.t. permutations. The fact that the domain is $mathbb R$ implies that you can lift this to a continuous function $beta:mathbb Rtomathbb R^n$ representing the roots
$endgroup$
– Federico
Nov 27 '18 at 18:27
$begingroup$
Basically, you have troubles only when multiple roots coincide. When they separate again (if they do), you just decide arbitrarily which $beta_i$ tracks which root
$endgroup$
– Federico
Nov 27 '18 at 18:30
|
show 16 more comments
1
$begingroup$
The functions $beta_i$ will not be polynomials in general. Consider for instance $p(x,t)=x^2+t$.
$endgroup$
– Federico
Nov 27 '18 at 18:03
$begingroup$
Moreover, see math.stackexchange.com/questions/940653/…
$endgroup$
– Federico
Nov 27 '18 at 18:05
$begingroup$
@Federico: Thanks. I see the functions could not be polynomial. But do they exist? The link was considering the domain $mathbb C$ whereas here the domain of interest is $mathbb R$.
$endgroup$
– user1101010
Nov 27 '18 at 18:26
$begingroup$
The roots of $p(,cdot,,t)$ are a continuous function $mathbb Rto mathbb C^n/mathrm{perm}$, where the quotient is w.r.t. permutations. The fact that the domain is $mathbb R$ implies that you can lift this to a continuous function $beta:mathbb Rtomathbb R^n$ representing the roots
$endgroup$
– Federico
Nov 27 '18 at 18:27
$begingroup$
Basically, you have troubles only when multiple roots coincide. When they separate again (if they do), you just decide arbitrarily which $beta_i$ tracks which root
$endgroup$
– Federico
Nov 27 '18 at 18:30
1
1
$begingroup$
The functions $beta_i$ will not be polynomials in general. Consider for instance $p(x,t)=x^2+t$.
$endgroup$
– Federico
Nov 27 '18 at 18:03
$begingroup$
The functions $beta_i$ will not be polynomials in general. Consider for instance $p(x,t)=x^2+t$.
$endgroup$
– Federico
Nov 27 '18 at 18:03
$begingroup$
Moreover, see math.stackexchange.com/questions/940653/…
$endgroup$
– Federico
Nov 27 '18 at 18:05
$begingroup$
Moreover, see math.stackexchange.com/questions/940653/…
$endgroup$
– Federico
Nov 27 '18 at 18:05
$begingroup$
@Federico: Thanks. I see the functions could not be polynomial. But do they exist? The link was considering the domain $mathbb C$ whereas here the domain of interest is $mathbb R$.
$endgroup$
– user1101010
Nov 27 '18 at 18:26
$begingroup$
@Federico: Thanks. I see the functions could not be polynomial. But do they exist? The link was considering the domain $mathbb C$ whereas here the domain of interest is $mathbb R$.
$endgroup$
– user1101010
Nov 27 '18 at 18:26
$begingroup$
The roots of $p(,cdot,,t)$ are a continuous function $mathbb Rto mathbb C^n/mathrm{perm}$, where the quotient is w.r.t. permutations. The fact that the domain is $mathbb R$ implies that you can lift this to a continuous function $beta:mathbb Rtomathbb R^n$ representing the roots
$endgroup$
– Federico
Nov 27 '18 at 18:27
$begingroup$
The roots of $p(,cdot,,t)$ are a continuous function $mathbb Rto mathbb C^n/mathrm{perm}$, where the quotient is w.r.t. permutations. The fact that the domain is $mathbb R$ implies that you can lift this to a continuous function $beta:mathbb Rtomathbb R^n$ representing the roots
$endgroup$
– Federico
Nov 27 '18 at 18:27
$begingroup$
Basically, you have troubles only when multiple roots coincide. When they separate again (if they do), you just decide arbitrarily which $beta_i$ tracks which root
$endgroup$
– Federico
Nov 27 '18 at 18:30
$begingroup$
Basically, you have troubles only when multiple roots coincide. When they separate again (if they do), you just decide arbitrarily which $beta_i$ tracks which root
$endgroup$
– Federico
Nov 27 '18 at 18:30
|
show 16 more comments
1 Answer
1
active
oldest
votes
$begingroup$
If all the coefficients of a polynomial are continuous functions over a real interval, then there is a selection of continuous functions that constitute the roots of the polynomial.
This is a result of Kato (as far as I know; it may be traced back to earlier publications.)
However, if the coefficients of a polynomial are continuous functions over a domain that contains the origin in the complex plane, then roots each as a continuous function do not exist in general. Note that roots are always continuous in the sense of topology. (F. Zhang)
answered Jan 29 at 17:22
Fuzhen ZhangFuzhen Zhang
111
$endgroup$
$begingroup$
Unless you provide a reference, I don't think that this shoud be considered an answer. And what's the meaning of thae last sentence?
$endgroup$
– José Carlos Santos
Jan 29 at 17:57
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If all the coefficients of a polynomial are continuous functions over a real interval, then there is a selection of continuous functions that constitute the roots of the polynomial.
This is a result of Kato (as far as I know; it may be traced back to earlier publications.)
However, if the coefficients of a polynomial are continuous functions over a domain that contains the origin in the complex plane, then roots each as a continuous function do not exist in general. Note that roots are always continuous in the sense of topology. (F. Zhang)
answered Jan 29 at 17:22
Fuzhen ZhangFuzhen Zhang
111
$endgroup$
$begingroup$
Unless you provide a reference, I don't think that this shoud be considered an answer. And what's the meaning of thae last sentence?
$endgroup$
– José Carlos Santos
Jan 29 at 17:57
add a comment |
$begingroup$
If all the coefficients of a polynomial are continuous functions over a real interval, then there is a selection of continuous functions that constitute the roots of the polynomial.
This is a result of Kato (as far as I know; it may be traced back to earlier publications.)
However, if the coefficients of a polynomial are continuous functions over a domain that contains the origin in the complex plane, then roots each as a continuous function do not exist in general. Note that roots are always continuous in the sense of topology. (F. Zhang)
answered Jan 29 at 17:22
Fuzhen ZhangFuzhen Zhang
111
$endgroup$
$begingroup$
Unless you provide a reference, I don't think that this shoud be considered an answer. And what's the meaning of thae last sentence?
$endgroup$
– José Carlos Santos
Jan 29 at 17:57
add a comment |
$begingroup$
If all the coefficients of a polynomial are continuous functions over a real interval, then there is a selection of continuous functions that constitute the roots of the polynomial.
This is a result of Kato (as far as I know; it may be traced back to earlier publications.)
However, if the coefficients of a polynomial are continuous functions over a domain that contains the origin in the complex plane, then roots each as a continuous function do not exist in general. Note that roots are always continuous in the sense of topology. (F. Zhang)
answered Jan 29 at 17:22
Fuzhen ZhangFuzhen Zhang
111
$endgroup$
If all the coefficients of a polynomial are continuous functions over a real interval, then there is a selection of continuous functions that constitute the roots of the polynomial.
This is a result of Kato (as far as I know; it may be traced back to earlier publications.)
However, if the coefficients of a polynomial are continuous functions over a domain that contains the origin in the complex plane, then roots each as a continuous function do not exist in general. Note that roots are always continuous in the sense of topology. (F. Zhang)
answered Jan 29 at 17:22
Fuzhen ZhangFuzhen Zhang
111
answered Jan 29 at 17:22
Fuzhen ZhangFuzhen Zhang
111
answered Jan 29 at 17:22
Fuzhen ZhangFuzhen Zhang
111
answered Jan 29 at 17:22
Fuzhen ZhangFuzhen Zhang
111
111
$begingroup$
Unless you provide a reference, I don't think that this shoud be considered an answer. And what's the meaning of thae last sentence?
$endgroup$
– José Carlos Santos
Jan 29 at 17:57
add a comment |
$begingroup$
Unless you provide a reference, I don't think that this shoud be considered an answer. And what's the meaning of thae last sentence?
$endgroup$
– José Carlos Santos
Jan 29 at 17:57
$begingroup$
Unless you provide a reference, I don't think that this shoud be considered an answer. And what's the meaning of thae last sentence?
$endgroup$
– José Carlos Santos
Jan 29 at 17:57
$begingroup$
Unless you provide a reference, I don't think that this shoud be considered an answer. And what's the meaning of thae last sentence?
$endgroup$
– José Carlos Santos
Jan 29 at 17:57
add a comment |
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1
$begingroup$
The functions $beta_i$ will not be polynomials in general. Consider for instance $p(x,t)=x^2+t$.
$endgroup$
– Federico
Nov 27 '18 at 18:03
$begingroup$
Moreover, see math.stackexchange.com/questions/940653/…
$endgroup$
– Federico
Nov 27 '18 at 18:05
$begingroup$
@Federico: Thanks. I see the functions could not be polynomial. But do they exist? The link was considering the domain $mathbb C$ whereas here the domain of interest is $mathbb R$.
$endgroup$
– user1101010
Nov 27 '18 at 18:26
$begingroup$
The roots of $p(,cdot,,t)$ are a continuous function $mathbb Rto mathbb C^n/mathrm{perm}$, where the quotient is w.r.t. permutations. The fact that the domain is $mathbb R$ implies that you can lift this to a continuous function $beta:mathbb Rtomathbb R^n$ representing the roots
$endgroup$
– Federico
Nov 27 '18 at 18:27
$begingroup$
Basically, you have troubles only when multiple roots coincide. When they separate again (if they do), you just decide arbitrarily which $beta_i$ tracks which root
$endgroup$
– Federico
Nov 27 '18 at 18:30