Zeros of an affine combination of two coprime polynomials

Question

Suppose $f, g \in \mathbb R[x]$ are two coprime monic polynomials with \begin{align*} f(x) = x^n + a_{n-1} x^{n-1} + \dots + a_0 \\ g(x) = x^m + b_{m-1} x^{m-1} + \dots + b_0, \end{align*} and $n > m$. Let $h(x,t) = f(x) - t g(x)$ where $t \in \mathbb R$. I read in a book without explanation that: If $t \in [0, \infty)$ and as $t \to \infty$, there are $n-m$ zeros of $h$ that would be asymptotically close to $n-m$ rays passing through $z = -\frac{a_{n-1}-b_{m-1}}{n-m}$. As $t \to \infty$, the $n-m$ zeros are arbitrarily close to \begin{align*} z = -\frac{a_{n-1} - b_{m-1} }{n-m} + t^{1/(n-m)}\omega_j, \end{align*} where $\omega_j$'s are the $(n-m)^{\text{th}}$ roots of unity. How do we formally prove this argument?

I can only $n-m$ zeros going to infinity by Rouche's Theorem. We can compare $tg$ and $f-tg$ on a sphere that encloses all the zeros of $g$. But have not idea on where the asymptotes come from.

Answer 1

Note first that if we substitute $x+c$ for $x$ in our polynomials, $a_{n-1}$ increases by $nc$ and $b_{m-1}$ increases by $mc$ and so $-\frac{a_{n-1}-b_{m-1}}{n-m}$ decreases by $c$ and the truth of the statement does not change. In particular, taking $c=-b_{m-1}/m$, we may assume $b_{m-1}=0$.

Now let $d=-\frac{a_{n-1}}{n-m}$, and let $f_0(x)=f(x)-x^m(x-d)^{n-m}$ and let $g_0(x)=g(x)-x^m$. Note that $\deg f_0\leq n-2$ and $\deg g_0\leq m-2$. We have $$h(x,t)=x^m((x-d)^{n-m}-t)+f_0(x)-tg_0(x).$$ Let $r=d+t^{1/(n-m)}\omega$ where $\omega$ is an $(n-m)$th root of unity and observe that $r$ is a root of $H(x,t)=x^m((x-d)^{n-m}-t)$. The idea is now to show that $H(x,t)$ dominates $f_0(x)-tg_0(x)$ as $x$ goes around a loop near $r$, so that by Rouche's theorem $h(x,t)$ will have to also have a root near $r$.

Fix $\epsilon>0$. We see that if $|x-r|=\epsilon$ then $|x^m|$ is within a constant factor of $t^{m/(n-m)}$ for $t$ sufficiently large. Also, $$\frac{(x-d)^{n-m}-t}{x-r}=\frac{(t^{1/(n-m)}\omega+(x-r))^{n-m}-(t^{1/(n-m)}\omega)^{n-m}}{x-r}$$ is equal to the derivative of $z\mapsto z^{n-m}$ at some point within $\epsilon$ of $t^{1/(n-m)}\omega$ by the mean value theorem. It follows that $|(x-d)^{n-m}-t|$ is within a constant factor of $t^{(n-m-1)/(n-m)}$ for $t$ sufficiently large. We thus see that for $|x-r|<\epsilon$, $|H(x,t)|$ is bounded below by a constant multiple of $t^{(n-1)/(n-m)}$ for $t$ sufficiently large.

On the other hand, for $|x-r|=\epsilon$, $|f_0(x)|$ is bounded above by a constant multiple of $t^{(n-2)/(n-m)}$ and $|g_0(x)|$ is bounded above by a constant multiple of $t^{(m-2)/(n-m)}$. It follows that $|f_0(x)-tg_0(x)|$ is bounded above by a constant multiple of $t^{(n-2)/(n-m)}$.

Thus, we see that if $t$ is sufficiently large, then $|H(x,t)|>|f_0(x)-tg_0(x)|$ for all $x$ such that $|x-r|=\epsilon$. So, by Rouche's theorem, since $H(x,t)$ has a root at $x=r$, we conclude that $h(x,t)=H(x,t)+f_0(x)-tg_0(x)$ must also have a root in the disk $|x-r|\leq\epsilon$.

Answer 2

To avoid the roots set $t=s^{n-m}$. Now scale the problem by setting $x=su$. Then $$ s^{-n}h(su, s^{n-m})=(u^n+s^{-1}a_{n-1}u^{n-1}+...+s^{-n}a_0)-(u^m+s^{-1}b_{m-1}u^{m-1}+...+s^{-m}b_0).$$ For $s\to\infty\iff t\to\infty$ the remaining polynomial $u^n-u^m$ has an $m$- fold root at $u=0$ and the other roots at the $(m-n)$th unit roots. For $s\approx\infty$ very large or $s^{-1}$ very small, the perturbation due to the next terms in $$ 0=u^m(u^{n-m}-1)+s^{-1}u^{m-1}(a_{n-1}u^{n-m}-b_{m-1})+O(s^{-2}u^{m-2}) $$ gives $m$ very small roots of size $O(s^{-1})$ close to the roots of $0=g(su)$. From each of the unperturbed unit roots $u=ω^k$, $ω^{n-m}=1$, the perturbation will result in a root $u=ω^k(1+s^{-1}v)+O(s^{-2})$, where $v$ solves $$ 0=ω^{mk}(n-m)v+ω^{(m-1)k}(a_{n-1}-b_{m-1})\implies v=-ω^{-k}\frac{a_{n-1}-b_{m-1}}{n-m} $$ so that in total $$ u=ω^k-s^{-1}\frac{a_{n-1}-b_{m-1}}{n-m}+O(s^{-2})~~\text{ or }~~ x=sω^k-\frac{a_{n-1}-b_{m-1}}{n-m}+O(s^{-1}) $$

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As an example I plotted the root locations for polynomials with $n=9,m=4$, $a_8=2$, $b_3=-3$, left for the $u$ parametrization, right for the $x$ parametrization. $s$ varies from $1$ to $50$ ($s^{-1}$ in equidistant steps). In light gray are the unit root locations resp. the asymptotic rays.