Polynomial Equations and Maximum Coefficients

If \( \alpha \) is a root of a polynomial equation \( P(x) = 0 \), then \( P(\alpha) = 0 \), and if \(M\) is the maximum coefficient value of \( |a_{n}|, |a_{n-1}|, |a_{n-2}|, \cdots ,|a_2|, |a_1| \) then \( nM \ge |a_{n}| + |a_{n-1}| + |a_{n-2}| + \cdots + |a_2| + |a_1| \). One of the inequalities of absolute value can be used effectively for proving polynomial equation properties, \( |x + y| \le |x| + |y| \). You can find the proof of absolute value inequalities.

Worked Example

Let \( \alpha \) be a root of \( P(x) = x^n +A_{n-1} x^{n-1} + A_{n-2} x^{n-2} + \cdots + A_2 x^2 + A_1 x + A_0 \).
Let \( M \) be the maximum value of \( |A_{n}|, |A_{n-1}|, |A_{n-2}|, \cdots , |A_2|, |A_1| \).

(a)    Show that \( |\alpha |^n \ge M(|\alpha|^{n-1} + |\alpha|^{n-2} + \cdots + |\alpha| + 1) \), given \( |a + b| \ge |a| + |b| \).

\( \begin{aligned} \require{color} \displaystyle
P(\alpha) &= 0 &\color{red} \text{since } \alpha \text{ is a root} \\
0 &= \alpha^n + A_{n-1} \alpha^{n-1} + A_{n-2} \alpha^{n-2} + \cdots + A_1 \alpha + A_0 \\
\alpha^n &= -A_{n-1} \alpha^{n-1} – A_{n-2} \alpha^{n-2} – \cdots – A_1 \alpha – A_0 \\
|\alpha^n| &= |-A_{n-1} \alpha^{n-1} – A_{n-2} \alpha^{n-2} – \cdots – A_1 \alpha – A_0| \\
|\alpha|^n &= |A_{n-1} \alpha^{n-1} + A_{n-2} \alpha^{n-2} + \cdots + A_1 \alpha + A_0| \\
|\alpha|^n &\le |A_{n-1} \alpha^{n-1}| + |A_{n-2} \alpha^{n-2}| + \cdots + |A_1 \alpha| + |A_0| &\color{red} |x + y| \le |x| + |y|\\
&= |A_{n-1}||\alpha^{n-1}| + |A_{n-2}||\alpha^{n-2}| + \cdots + |A_1||\alpha| + |A_0| \\
&\le M|\alpha^{n-1}| + M|\alpha^{n-2}| + \cdots + M|\alpha| + M \\
&= M(|\alpha^{n-1}| + |\alpha^{n-2}| + \cdots + |\alpha| + 1) \\
&\le M(|\alpha|^{n-1} + |\alpha|^{n-2} + \cdots + |\alpha| + 1) &\color{red} |\alpha^n| \le |\alpha|^n \\
\therefore |\alpha|^n &\le M(|\alpha|^{n-1} + |\alpha|^{n-2} + \cdots + |\alpha| + 1) \\
\end{aligned} \\ \)

(b)    Show that \(1 < | \alpha| < 1 + M \) for any root \( \alpha \) of \( P(x) \).

\( \begin{aligned} \displaystyle \require{color}\ \newcommand\ddfrac[2]{\frac{\displaystyle #1}{\displaystyle #2}}
|\alpha|^n &\le (|\alpha|^{n-1} + |\alpha|^{n-2} + \cdots + |\alpha| + 1) \\
|\alpha|^n &\le M \times \frac{|\alpha|^n – 1}{|\alpha| – 1} &\color{red} a + ar + ar^2 + \cdots + ar^{n-1} = \frac{a(r^2 – 1)}{r-1} \\
1 &\le M \times \frac{1- \ddfrac{1}{|\alpha|^n}}{|\alpha|-1} &\color{red}\text{divide both sides by } |\alpha|^n \\
1 &\le M \times \frac{1- \ddfrac{1}{|\alpha|^n}}{|\alpha|-1} < \frac{1}{|\alpha|-1} \\
1 &< \frac{1}{|\alpha|-1} \\
(|\alpha|-1)^2 &< M \times (|\alpha|-1) &\color{red}\text{multipy both sides by } (|\alpha|-1)^2 \\
(|\alpha|-1)^2- M \times (|\alpha|-1) &<0 \\
(|\alpha|-1)( |\alpha| – 1- M ) &< 0 \\
\therefore 1 &< |\alpha| < 1 + M \\
\end{aligned} \)

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