Showing that the roots of a polynomial with descending positive coefficients lie in the unit disc.

Question

Let $P(z)=a_0+a_1z+\cdots+a_nz^n$ be a polynomial whose coefficients satisfy $$0<a_0<a_1<\cdots<a_n.$$

I want to show that the roots of $P$ live in unit disc. The obvious idea is to use Rouche's theorem, but that doesn't quite work here, at least with the choice $f(z)=a_nz^n, g(z)=$ (the rest).

Any ideas?

Answer 1

The thing to do is to look instead at the polynomial $$Q(z) = (1-z)P(z) = (1-z)\left(\sum_{i=0}^n a_iz^i \right) = a_0 -a_n z^{n+1} + \sum_{i=1}^n (a_i-a_{i-1})z^i$$ Now, let $|z|>1$ be a root of $P(z)$, and hence a root of $Q(z)$. Therefore, we have $a_0 + \sum_{i=1}^n (a_i-a_{i-1})z^i = a_n z^{n+1}$ Then, we have \begin{aligned} |a_n z^{n+1}| &= \left|a_0 + \sum_{i=1}^n (a_i-a_{i-1})z^i\right| \\ & \le a_0 + \sum_{i=1}^n (a_i-a_{i-1})|z^i| \\ & < a_0|z^n| + \sum_{i=1}^n (a_i-a_{i-1})|z^n| \\ & = |a_n z^n|\end{aligned} a contradiction.

For a nice article on integer polynomials, see here. (Your problem is Proposition 10)

Answer 2

The previous answer establishes that roots lie inside or on the unit circle. By a slight modification to $$ Q(z)=(1-z)P(Rz) $$ this argument remains true as long as $0<a_0\le Ra_1 \le R^2a_2\le ... \le R^na_n$. Let $R$ be minimal with this property, then $R<1$. Now if $z$ is any root of $P$, then $z/R$ is a root of $Q$ and thus in the unit disk. Thus $|z|\le R<1$.

Answer 3

To simplify notation we can assume WLOG that $c_0=1$
(a) $p(z)=c_n z^n + \dots + c_1 z + 1$
(b) $q(z) := z^n + \dots + c_{n-1}z + c_n \text{ (reciprocal polynomial)}$

(c) Modified Schur Transform
$f(z) := c_0\cdot p(z) - \delta\cdot c_n\cdot q(z)= p(z) - \frac{1}{c_n}\cdot q(z)=\sum_{k=1}^n \big(c_k-\frac{c_{n-k}}{c_n}\big)\cdot z^k$ $\text{degree }f=n \text{ and } f(0)=0 \text{ and }\frac{f(z)}{z} \text{ has positive increasing coefficients since}$
$\frac{c_{n-1}}{c_n}\lt 1 =c_0\lt c_1 \text{ and for } 1\leq k\leq n-1\text{: }$
$c_k-\frac{c_{n-k}}{c_n} \lt c_{k+1}-\frac{c_{n-(k+1)}}{c_n} $ which holds since $c_k \lt c_{k+1} \text{ and }-\frac{c_{n-k}}{c_n}\lt-\frac{c_{n-k-1}}{c_n} $

$\implies \frac{f(z)}{z} $ has all $n-1$ roots in $\mathbb D$, the (open) unit disc, by induction hypothesis. (The Base Case trivially holds for degree one polynomials with descending positive coefficients -- their sole root is given by $\frac{-c_0}{c_1}\in \mathbb D$.)

$\implies f(z) $ has all $n$ roots in $\mathbb D\implies p(z) $ has all $n$ roots in $\mathbb D$
the second implication holds because $\big\vert p(w)\big\vert \gt \big \vert \frac{1}{c_n}q(w)\big \vert$ for $w \in \partial \mathbb D$ $\Big[$since $\big \vert p\big(w\big)\big\vert=\big \vert q\big(w\big)\big\vert\Big]$ so the result follows by Rouche. Note that $p(z)$ cannot have roots $\in\partial \mathbb D$ since $p(w)=0\implies f(w)=0\implies f(z)$ has (at least) $n+1$ roots but degree $n$ (impossible).