Root dependence on coefficient terms
$begingroup$
I'm trying to solve how to roots of this polynomial depend on $k$:
$$ x^3 + (c_2 + a_1 k^2) x^2 + (c_1 + a_2 k^2 + a_3 k^4) x + (c_0 + a_4 k^2 + a_5 k^4 + a_6 k^6) = 0 $$
I could put this into the cubic formula, and I've tried doing this, but it's a brutal slog. I'm thinking of a general strategy of I was wondering if there are any general properties of polynomial roots that could help me with this? Or is this problem just analytically intractable?
For what it's worth, I'm only interested in $k>1$.
polynomials
asked Jan 20 at 23:52
Mike FlynnMike Flynn
6271617
$endgroup$
add a comment |
$begingroup$
I'm trying to solve how to roots of this polynomial depend on $k$:
$$ x^3 + (c_2 + a_1 k^2) x^2 + (c_1 + a_2 k^2 + a_3 k^4) x + (c_0 + a_4 k^2 + a_5 k^4 + a_6 k^6) = 0 $$
I could put this into the cubic formula, and I've tried doing this, but it's a brutal slog. I'm thinking of a general strategy of I was wondering if there are any general properties of polynomial roots that could help me with this? Or is this problem just analytically intractable?
For what it's worth, I'm only interested in $k>1$.
polynomials
asked Jan 20 at 23:52
Mike FlynnMike Flynn
6271617
$endgroup$
add a comment |
$begingroup$
I'm trying to solve how to roots of this polynomial depend on $k$:
$$ x^3 + (c_2 + a_1 k^2) x^2 + (c_1 + a_2 k^2 + a_3 k^4) x + (c_0 + a_4 k^2 + a_5 k^4 + a_6 k^6) = 0 $$
I could put this into the cubic formula, and I've tried doing this, but it's a brutal slog. I'm thinking of a general strategy of I was wondering if there are any general properties of polynomial roots that could help me with this? Or is this problem just analytically intractable?
For what it's worth, I'm only interested in $k>1$.
polynomials
asked Jan 20 at 23:52
Mike FlynnMike Flynn
6271617
$endgroup$
I'm trying to solve how to roots of this polynomial depend on $k$:
$$ x^3 + (c_2 + a_1 k^2) x^2 + (c_1 + a_2 k^2 + a_3 k^4) x + (c_0 + a_4 k^2 + a_5 k^4 + a_6 k^6) = 0 $$
I could put this into the cubic formula, and I've tried doing this, but it's a brutal slog. I'm thinking of a general strategy of I was wondering if there are any general properties of polynomial roots that could help me with this? Or is this problem just analytically intractable?
For what it's worth, I'm only interested in $k>1$.
polynomials
polynomials
asked Jan 20 at 23:52
Mike FlynnMike Flynn
6271617
asked Jan 20 at 23:52
Mike FlynnMike Flynn
6271617
asked Jan 20 at 23:52
Mike FlynnMike Flynn
6271617
asked Jan 20 at 23:52
Mike FlynnMike Flynn
6271617
asked Jan 20 at 23:52
Mike FlynnMike Flynn
6271617
6271617
add a comment |
add a comment |
1 Answer
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$begingroup$
If $kgg1$ it might be worthwhile to substitute
$$x:= k^2,y,qquad{1over k^2}=:epsilonll1 .$$
In this way we arrive at the equation
$$y^3+(a_1+c_2epsilon)y^2+(a_3+a_2epsilon+c_1epsilon^2)y+(a_6+a_5epsilon+a_4epsilon^2+c_0epsilon^3)=0 .tag{1}$$
You first have to solve
$$y^3+a_1y^2+a_3y+a_6=0$$
numerically (or exactly). Assume that $etain{mathbb C}$ is one of the solutions. You can then plug the "Ansatz"
$$y(epsilon)=eta+p_1epsilon+p_2epsilon^2+ldots$$
into $(1)$ and obtain linear equations for $p_1$, $p_2$, $ldots $. In this way the $p_k$ become expressions in terms of $eta$ and the $a_j$, $c_j$.
answered Jan 21 at 10:16
Christian BlatterChristian Blatter
174k8115327
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If $kgg1$ it might be worthwhile to substitute
$$x:= k^2,y,qquad{1over k^2}=:epsilonll1 .$$
In this way we arrive at the equation
$$y^3+(a_1+c_2epsilon)y^2+(a_3+a_2epsilon+c_1epsilon^2)y+(a_6+a_5epsilon+a_4epsilon^2+c_0epsilon^3)=0 .tag{1}$$
You first have to solve
$$y^3+a_1y^2+a_3y+a_6=0$$
numerically (or exactly). Assume that $etain{mathbb C}$ is one of the solutions. You can then plug the "Ansatz"
$$y(epsilon)=eta+p_1epsilon+p_2epsilon^2+ldots$$
into $(1)$ and obtain linear equations for $p_1$, $p_2$, $ldots $. In this way the $p_k$ become expressions in terms of $eta$ and the $a_j$, $c_j$.
answered Jan 21 at 10:16
Christian BlatterChristian Blatter
174k8115327
$endgroup$
add a comment |
$begingroup$
If $kgg1$ it might be worthwhile to substitute
$$x:= k^2,y,qquad{1over k^2}=:epsilonll1 .$$
In this way we arrive at the equation
$$y^3+(a_1+c_2epsilon)y^2+(a_3+a_2epsilon+c_1epsilon^2)y+(a_6+a_5epsilon+a_4epsilon^2+c_0epsilon^3)=0 .tag{1}$$
You first have to solve
$$y^3+a_1y^2+a_3y+a_6=0$$
numerically (or exactly). Assume that $etain{mathbb C}$ is one of the solutions. You can then plug the "Ansatz"
$$y(epsilon)=eta+p_1epsilon+p_2epsilon^2+ldots$$
into $(1)$ and obtain linear equations for $p_1$, $p_2$, $ldots $. In this way the $p_k$ become expressions in terms of $eta$ and the $a_j$, $c_j$.
answered Jan 21 at 10:16
Christian BlatterChristian Blatter
174k8115327
$endgroup$
add a comment |
$begingroup$
If $kgg1$ it might be worthwhile to substitute
$$x:= k^2,y,qquad{1over k^2}=:epsilonll1 .$$
In this way we arrive at the equation
$$y^3+(a_1+c_2epsilon)y^2+(a_3+a_2epsilon+c_1epsilon^2)y+(a_6+a_5epsilon+a_4epsilon^2+c_0epsilon^3)=0 .tag{1}$$
You first have to solve
$$y^3+a_1y^2+a_3y+a_6=0$$
numerically (or exactly). Assume that $etain{mathbb C}$ is one of the solutions. You can then plug the "Ansatz"
$$y(epsilon)=eta+p_1epsilon+p_2epsilon^2+ldots$$
into $(1)$ and obtain linear equations for $p_1$, $p_2$, $ldots $. In this way the $p_k$ become expressions in terms of $eta$ and the $a_j$, $c_j$.
answered Jan 21 at 10:16
Christian BlatterChristian Blatter
174k8115327
$endgroup$
If $kgg1$ it might be worthwhile to substitute
$$x:= k^2,y,qquad{1over k^2}=:epsilonll1 .$$
In this way we arrive at the equation
$$y^3+(a_1+c_2epsilon)y^2+(a_3+a_2epsilon+c_1epsilon^2)y+(a_6+a_5epsilon+a_4epsilon^2+c_0epsilon^3)=0 .tag{1}$$
You first have to solve
$$y^3+a_1y^2+a_3y+a_6=0$$
numerically (or exactly). Assume that $etain{mathbb C}$ is one of the solutions. You can then plug the "Ansatz"
$$y(epsilon)=eta+p_1epsilon+p_2epsilon^2+ldots$$
into $(1)$ and obtain linear equations for $p_1$, $p_2$, $ldots $. In this way the $p_k$ become expressions in terms of $eta$ and the $a_j$, $c_j$.
answered Jan 21 at 10:16
Christian BlatterChristian Blatter
174k8115327
answered Jan 21 at 10:16
Christian BlatterChristian Blatter
174k8115327
answered Jan 21 at 10:16
Christian BlatterChristian Blatter
174k8115327
answered Jan 21 at 10:16
Christian BlatterChristian Blatter
174k8115327
174k8115327
add a comment |
add a comment |
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