Why is $(1+B/A)$ a superior bound of the module of the roots of a polynomial?

Question

Let $A$ be the coefficient of the term of greatest degree of a polynomial and $B$ be the maximum of the modules of the other coefficients.

How can I prove that $(1+B/A)$ is a superior bound of the module of the roots of the polynomial?

Answer 1

Consider a degree $n$ polynomial $p(x) = \sum_{i=0}^n c_i x^i$.

Let $q(x)=\frac{p(x)}{c_n}=\sum_{i=0}^n \frac{c_i}{c_n}x^i.$ $q(x)$ and $p(x)$ would share the same roots.

The companion matrix of $q(x)$ is $\begin{bmatrix} 0 & 0 & \ldots & 0 & -\frac{c_0}{c_n} \\ 1 & 0 & \ldots & 0 & -\frac{c_1}{c_n} \\ 0 & 1 & \ldots & 0 & -\frac{c_2}{c_n} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \ldots & 1 & - \frac{c_{n-1}}{c_n}\end{bmatrix}.$

Let $\lambda$ be a root, By Gershgrorin's theorem,

$$\left(|\lambda| \le \frac{|c_0|}{A} \right)\lor \left(|\lambda| \le 1+\frac{\max_{1\le i \le n-2}|c_i|}{A}\right) \lor \left( \left| \lambda+ \frac{c_{n-1}}{c_n}\right| \le 1 \right) $$

$$\implies \left(|\lambda| \le 1+\frac{\max_{0\le i \le n-2}|c_i|}{A}\right) \lor \left( -\frac{c_{n-1}}{c_n}-1 \le \lambda \le -\frac{c_{n-1}}{c_n}+1\right)$$

$$\implies \left(|\lambda| \le 1+\frac{\max_{0\le i \le n-2}|c_i|}{A}\right) \lor \left( | \lambda| \le 1+\frac{|c_{n-1}|}{A}\right)$$

$$\implies |\lambda| \le 1+\frac{\max_{0\le i \le n-1}|c_i|}{A} $$