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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 $k>1$.

Mike Flynn
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1 Answers1

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If $k\gg1$ it might be worthwhile to substitute $$x:= k^2\,y,\qquad{1\over k^2}=:\epsilon\ll1\ .$$ In this way we arrive at the equation $$y^3+(a_1+c_2\epsilon)y^2+(a_3+a_2\epsilon+c_1\epsilon^2)y+(a_6+a_5\epsilon+a_4\epsilon^2+c_0\epsilon^3)=0\ .\tag{1}$$ You first have to solve $$y^3+a_1y^2+a_3y+a_6=0$$ numerically (or exactly). Assume that $\eta\in{\mathbb C}$ is one of the solutions. You can then plug the "Ansatz" $$y(\epsilon)=\eta+p_1\epsilon+p_2\epsilon^2+\ldots$$ 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 $a_j$, $c_j$.

Christian Blatter
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