Bounds on roots of polynomials

Question

What bound is there on the roots of a given polynomial, in terms of its degree and coefficients?

Consider the polynomial $p(x) = 3x^7 – 5x^3 + 42$. Would you not agree, without doing any calculation, that one million ($10^6$) cannot be a root? It just wouldn’t be in accord with the smallness of the coefficients and the well-behavedness of polynomials. And yet I don’t recall ever having encountered anything in the literature that gave a bound on the absolute value of the roots of a polynomial in terms of the degree and coefficients of the polynomial, but I’m pretty sure such must exist, and that I simply missed it, and so I’m tagging this a a reference-request.

By the way, during my post of a question, every time, it seems, there are stray graphics on the screen, as if from someone else's question or answer. Is this happening to anyone else?

Answer 1

Here's an elementary bound. Let $p(x) = x^n - a_{n-1} x^{n-1} - ... - a_0$. If $|x| \ge 1$ then $m \ge n$ implies $|x|^m \ge |x|^n$, hence if $p(x) = 0$ then

$$|x|^n = \left| a_{n-1} x^{n-1} + ... + a_0 \right| \le |x|^{n-1} \left( |a_{n-1}| + ... + |a_0| \right)$$

hence

$$|x| \le \text{max}(1, |a_{n-1}| + ... + |a_0|).$$

In your case we can do much better because of the zero coefficients: we get

$$|x|^7 \le \text{max}(1, \frac{47}{3} |x|^3)$$

hence $|x|^4 \le \frac{47}{3} < 16$, or $|x| < 2$.

A technique useful in special cases is the Gauss-Lucas theorem, and a technique useful in general cases is Rouche's theorem.

Answer 2

If $a_n x^n + \dots + a_1 x + a_0 = 0$ we get the following useful bound by Lagrange:

$$|x| \le 2\max\left\{ \left(\frac{|a_{i}|}{|a_n|}\right)^{\frac1{n-i}} : 0 \le i < n \right\}.$$

The following proof by Chee-Keng Yap is simple: Let $\beta=\max\left\{\left(|a_{i}|/|a_n|\right)^{1/(n-i)} : 0 \le i < n\right\}$. The lemma is trivial if $|x| \le \beta$; so we focus on $|x| > \beta$.

Since $a_n x^n = -(a_{n-1}x^{n-1}+\dots+a_0)$ we have $|a_n| |x|^n \le |a_{n-1}||x|^{n-1}+\dots+|a_0|$ by the triangle inequality. Hence $$ 1 \le \frac{1}{|a_n||x|^n}\sum_{i=1}^{n} |a_{n-i}||x^{n-i}| = \sum_{i=1}^{n}{\frac{|a_{n-i}|}{|a_n||x|^i}} \le \sum_{i=1}^{n}{\frac{\beta^{i}}{|x|^{i}}}\le \frac{\beta/|x|}{1-\beta/|x|}, $$ where we bounded the sum by the infinite geometric sum, which is finite since we assumed $\beta < |x|$. Now, $1\le \frac{\beta/|x|}{1-\beta/|x|}$ implies $1\le 2\beta/|x|$, which is what we want.

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In your case this says $$ |x| \le 2 \max\{\sqrt[4]{5/3}, \sqrt[7]{42/3}\} \approx 2 \max\{1.13622, 1.45792\} < 3. $$ So not quite as strong as Qiaochu's bound in this case, but in other cases it can be much better.

You can also try some more bounds from this Wikipedia article.