One of zeros of convex combination of polynomials goes to infinity?

Question

Suppose we have two polynomials of degree $n$ and $n-1$, \begin{align*} f(x) &= a_n x^n + a_{n-1} x^{n-1} + \dots + a_0, \\ g(x) &= b_{n-1}x^{n-1} + b_{n-2} x^{n-2} + \dots + b_0, \end{align*} with $a_{n} \neq 0$ and $b_{n-1} \neq 0$. Let us consider the convex combination of the two polynomials $h(x,t) = (1-t) f(x) + t g(x)$. I read a statement that says: as $t \to 1$, one of the zeros $h(x,t)$ would go to infinity. Intuitively, I think this is true because by Vieta's formula, as $t \to 1$, the coefficient with $x^n$ would go to $0$ and it must be that one of the root should be unbounded. How do we argue this point?

Answer 1

It is a consequence of Rouche's theorem. We can prove that for all sufficiently large $R>0$, there exists $\delta>0$ such that for each $t$ with $0<|t|<\delta$, there exists unique $z=z(t)$ such that $$ |z|\geq R, \;tf(z) + (1-t)g(z) =0. $$ Without loss of generality, let $$ R>\max\{|\lambda_1|,| \lambda_2|,\ldots, |\lambda_{n-1}|\},$$ where $\lambda_1, \lambda_2,\ldots, \lambda_{n-1}$ are roots of $g(\lambda)=0$ (counted with multiplicity.) And let $h_t(z)= g(z) +t(f(z)-g(z)).$ Now, consider $$ N:t\mapsto \frac{1}{2\pi i}\int_{|z|=R} \frac{h_t'(z)}{h_t(z)}dz. $$ If $h_t$ does not vanish on $|z|=R$, then $N$ gives us the number of roots of $h_t=0$ in the region $|z|<R$ by Cauchy's argument principle. Note that if $$ |t|<\delta:=\frac{\min_{|z|=R} |g(z)|}{\max_{|z|=R}|f(z)-g(z)|} \leq \min_{|z|=R}\frac{|g(z)|}{|f(z)-g(z)|}, $$ then $h_t$ does not vanish on $|z|=R$, and $N:t\in (-\delta,\delta)\to \mathbb{N}\cup \{0\}$ defines a continuous integer-valued function. Since it holds that $N(0)= n-1$ by the choice of $R$, we have $$N \equiv n-1$$ on $|t|<\delta$. Note that if $t\neq 0$, then by fundamental theorem of algebra, $h_t$ has $n$-roots counted with multiplicity. This implies that for all $0<|t|<\delta$, there exists exactly one root $z_t \in \mathbb{C}\setminus \mathbb{D}(0,R)$ of $h_t(z)=0$ as desired.

Answer 2

Consider the reverse polynomials \begin{align} \tilde h(w,t)&=w^nh(w^{-1},t)=(1−t)x^nf(w^{-1})+tx^ng(w^{-1})\\ &= \begin{array}{lll} &(1−t)&(a_n+&a_{n-1}w+...+a_1w^{n-1}+a_0w^n)\\ +&t~&(&b_{n-1}w+...+b_1w^{n-1}+b_0w^n) \end{array} \\ &=(1-t)\tilde f(w)+tw\tilde g(w) \end{align} Then one easily sees that $\tilde h(w,t)$ has always $n$ roots $w$. These correspond to the roots of $h(t,x)$ via $x=w^{-1}$ for $t<1$. At time $t=1$ the polynomial $\tilde h(w,1)=w\tilde g(w)$ has $w=0$ and the inverses of the roots of $g$ as roots. The set of roots of $\tilde h$ is a continuous function of $t$. As $\tilde h$ has a continuous path $w_0(t)$, $t\in [a,1]$ of roots converging to $0$, the original homotopy $h$ has a path $x_0(t)=w_0(t)^{-1}$, $t\in [a,1)$, of roots that diverges to infinity.