An upper bound for roots of polynomials

A natural question to ask when given a polynomial is: how do the roots and the coefficients interplay with one another? In fact a huge portion of polynomial theory and solving algebraic equations is finding out the details of this relationship. The quadratic formula is a prime example – It gives a way of finding roots by only using the values of the coefficients. An explicit relationship, an algebraic formula, is only possible for polynomials of degree less than 5 which is a famous result pioneered by the work of Galois.

The interplay we are going to determine for polynomials of any degree is that you only need to look for roots in a neighborhood around 0 with the neighborhood’s size depending only on the size of the polynomial’s coefficients.

Let p \in \mathbb{R}[x] be a polynomial over \mathbb{R} of degree n with the representation p(x) = a_nx^n +... a_1x + a_0 where a_n \neq 0 and that |a_i| \leq A for all i = 1,..., a and some A \in \mathbb{R} . We will be considering our polynomial as a function p : \mathbb{C} \longrightarrow \mathbb{C} and be setting M \geq 0 and R = 1 + \frac{A + M}{|a_n|} where our choice of R will be more apparent later.

What we are going to show is that for x of a large enough magnitude our polynomial will have a magnitude as large as we like. We will also establish a simple bound on the magnitude of the roots of our polynomial p .

Suppose |x| \geq R , we have |a_n x^n| = \frac{A + M}{R- 1}|x|^n by the definition of our constants and estimating |a_{n-1}x^{n-1} + .... + a_0|  gives | a_{n-1}x^{n-1} + .... + a_0 | \leq A (|x^{n-1}| + .... + |x| + 1 ) = A\frac{|x|^n - 1}{|x| - 1 } < A \frac{|x|^n}{R-1} by using the sum of a geometric series and the triangle inequality.

Using the triangle inequality allows us to conclude |a_nx^n| - |a_{n-1}x^{n-1} + ... + a_0 | \leq |p(x)| which gives us by using the previous estimation |p(x)| \geq \frac{A+ M}{R-1} |x|^n - \frac{A}{R-1} |x|^n = \frac{M|x|^n}{R-1} \geq \frac{MR^n}{R-1} > M as both |x| \leq R and R^n > R > R - 1 as R > 1.

Bringing this together we have shown for all M \geq 0 if we look at |x| \geq R where R = 1 + \frac{A + M}{|a_n|} we have |p(x)| > M .

Looking at the contrapositive of this statement lets us conclude if |p(x)| \leq M for some x \in \mathbb{C} and M \geq 0 we must have |x| < R or equivalently x living in the ball B(0;R). If we take x to be a root of p then by setting M = 0 we have that |x| < R = 1 +\frac{A}{|a_n|} giving us a bound on the size of our root in terms of the polynomials coefficients. This means all roots of p will be found in the ball B(0; a + \frac{A}{|a_n|} ).



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