﻿ Trapping the Roots of a Polynomial

# ‘TRAPPING’ THE ROOTS OF A POLYNOMIAL; AN INTERVAL GUARANTEED TO CONTAIN ALL REAL ROOTS

• PRACTICE (online exercises and printable worksheets)

It's often helpful to have an interval that is guaranteed to hold all the real roots of a polynomial.
This section states and proves such a theorem—it's a beautiful proof,
which uses some important properties of the real numbers that have not yet appeared in this course.

In particular, if you're using a calculator to graph a polynomial,
then you probably want to set your window so you're guaranteed to see all the $x$-intercepts.
This section is for you!

Here's the outline for this section:

Theorem:   Bounding the Real Roots of a Polynomial
an interval guaranteed to contain all real roots
 Let $\,P\,$ be a polynomial of degree $\,n\ge 1\,$ with real coefficients. That is, $$P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$$ where $\,a_n\ne 0\,$ and all the $\,a_i\,$ are real numbers. Let $\displaystyle\,M := \text{max}\bigl(\,|a_{n-1}|,\ldots,|a_0|\,\bigr)\,$. Let $\,r\,$ be any real root (zero) of $\,P\,$, so that $\,P(r) = 0\$.   Then: $$|r| \le 1 + \frac{M}{|a_n|}$$ If we define $\displaystyle\,K := 1 + \frac{M}{|a_n|}\,$, then: all the real zeros of $\,P\,$ are contained in the interval $\,[-K,K]\,$

## Notes on the theorem

• recall that ‘:=’ means ‘equals, by definition’
• to find $\,M\,$:   look at all the coefficients except the leading coefficient,
and choose the one with the biggest absolute value
• You may want to compare the theorem and proof on this page with Cauchy's Bound for the Roots of a Polynomial.
• The theorem and proof given on this page provides a bound for all complex roots of a polynomial.
In this case, $\,K\,$ is the radius of a bounding circle centered at the origin in the complex plane.
For a complex number $\,z\,$, $\,|z|\,$ is called the modulus of $\,z\,$.
$$\,|z| \ =\ |a + bi|\ =\ \text{distance of \,z\, from the origin in the complex plane} \ =\ \sqrt{a^2 + b^2}$$
• The theorem and proof given on this page work when the polynomial coefficients $\,a_i\,$ are complex numbers.

## Example 1

Let $\,P(x) = (x-1)(x+2)(x-3) = x^3 - 2x^2 - 5x + 6\,$.

It's clear from the factored form that the roots of $\,P\,$ are $\,1\,$, $\,-2\,$, and $\,3\,$,
so we'll be able to confirm the bounding interval obtained from the theorem.

$n\,$ (the degree) is $3$

$a_3\,$ (the leading coefficient) is $\,1\,$

$M = \text{max}\bigl(\,|-2|\,,\,|-5|\,,\,|6|\,\bigr) = 6$

$\displaystyle K \ =\ 1 + \frac{M}{|a_n|} \ =\ 1 + \frac{6}{|1|} \ =\ 1 + 6 = 7$

All the zeroes are guaranteed to lie in the interval $\,[-K,K] = [-7,7]\,$ (which they do!)

## Example 2

The previous example is tweaked by changing the leading coefficient to $\,7\,$.

Let $\,P(x) = 7x^3 - 2x^2 - 5x + 6\,$.

$n\,$ (the degree) is $3$

$a_3\,$ (the leading coefficient) is $\,7\,$

$M = \text{max}\bigl(\,|-2|\,,\,|-5|\,,\,|6|\,\bigr) = 6$

$\displaystyle K \ =\ 1 + \frac{M}{|a_n|} \ =\ 1 + \frac{6}{|7|} \ =\ \frac{13}{7}\ \approx\ 1.9$

All the zeroes are guaranteed to lie in the interval $\,[-\frac{13}{7},\frac{13}{7}]\,$.

A quick trip to WolframAlpha shows that there is only one real root, which is approximately $\,-1.1\,$.
Here's a sketch from WolframAlpha that shows all the complex roots: ## TOOLS NEEDED TO PROVE THE THEOREM

1. THE TRIANGLE INEQUALITY:
For all real numbers $\,a\,$ and $\,b\,$:   $|a+b| \le |a| + |b|$
2. PROPERTIES OF ABSOLUTE VALUE:
For all real numbers $\,a\,$ and $\,b\,$:
• $|ab| = |a|\cdot|b|$
• $\displaystyle \left|\frac{a}{b}\right| = \frac{|a|}{|b|}$     ($b\ne 0$)
3. Properties (1) and (2) extend to finite sums/products.
For example:
$|a + b + c| \le |a| + |b + c| \le |a| + |b| + |c|$
and
$|abc| = |a|\cdot|bc| = |a|\cdot|b|\cdot|c|$

In particular: $|x^n| = |x|^n$
4. Properties (1) and (2) also hold for complex numbers, where $\,|\cdot|\,$ is the modulus of a complex number.
Complex numbers are explored in the next few sections.
5. SUM OF A FINITE GEOMETRIC SERIES:
For all real numbers $\,a\,$ and $\,x\,$:
$\displaystyle a + ax + ax^2 + \cdots + ax^n = \frac{a(x^{n+1}-1)}{x-1}$

## PROOF OF THE BOUNDING THE REAL ROOTS OF A POLYNOMIAL THEOREM

Let $\,r\,$ be a real root of $\,P\,$, so $\,P(r) = 0\$: $$a_nr^n + a_{n-1}r^{n-1} + \cdots + a_1r + a_0 = 0$$ Dividing both sides of the equation by $\,a_n\ne 0\,$ gives the equivalent equation: $$r^n + \frac{a_{n-1}}{a_n}r^{n-1} + \cdots + \frac{a_1}{a_n}r + \frac{a_0}{a_n} = 0 \tag{1}$$

Define $\displaystyle\,M := \text{max}\bigl(\,|a_{n-1}|,\ldots,|a_0|\,\bigr)\,$, so that $$\text{max}\left(\left|\frac{a_{n-1}}{a_n}\right|,\ldots,\left|\frac{a_0}{a_n}\right|\right) = \text{max}\left(\frac{|a_{n-1}|}{|a_n|},\ldots,\frac{|a_0|}{|a_n|}\right) = \frac{\text{max}(|a_{n-1}|,\ldots,|a_0|)}{|a_n|} = \frac{M}{|a_n|}$$

We consider two cases: $|r|\le 1\,$ and $|r| > 1\,$.   In both cases, it must be shown that $\displaystyle\,|r| \le 1 + \frac{M}{|a_n|}\,$.

## $|r| \le 1$

Assume $\,|r| \le 1\,$. Since absolute value is a nonnegative quantity, $\displaystyle\,\frac{M}{|a_n|} \ge 0\,$.
Then, as desired, we have: $$|r| \quad\le\quad 1 \quad\le\quad 1 + \frac{M}{|a_n|}$$

## $|r| > 1$

Assume $|r| > 1$. Rewrite (1) as $$r^n = -\frac{a_0}{a_n} - \frac{a_1}{a_n}r - \cdots - \frac{a_{n-1}}{a_n}r^{n-1}\tag{2}$$ Then: \begin{alignat}{2} |r^n| \quad &= \quad \left|-\frac{a_0}{a_n} - \frac{a_1}{a_n}r - \cdots - \frac{a_{n-1}}{a_n}r^{n-1}\right| &&\cr\cr &\le \quad \left|\frac{a_0}{a_n}\right| + \left|\frac{a_1}{a_n}r\right| + \cdots + \left|\frac{a_{n-1}}{a_n}r^{n-1}\right| &\qquad\qquad&\text{triangle inequality}\cr\cr &= \quad \frac{|a_0|}{|a_n|} + \frac{|a_1|}{|a_n|}|r| + \cdots + \frac{|a_{n-1}|}{|a_n|}|r^{n-1}| &\qquad\qquad& \text{since} \left|\frac{a}{b}\right| = \frac{|a|}{|b|} \text{ and } |ab| = |a|\cdot|b|\cr\cr &= \quad \frac{1}{|a_n|}\left(|a_0| + |a_1|\cdot|r| + \cdots + |a_{n-1}|\cdot|r^{n-1}|\right) &&\text{factor out the common factor} \cr\cr &\le \quad \frac{1}{|a_n|}\left(M + M|r| + \cdots + M|r^{n-1}|\right) &&\text{since M is the max}\cr\cr &\le \quad \frac{1}{|a_n|}\left(M + M|r| + \cdots + M|r|^{n-1}\right) &&\text{since |r^i| = |r|^i}\cr\cr &= \quad \frac{M}{|a_n|}(1 + |r| + \cdots + |r|^{n-1}) &&\text{factor out M}\cr\cr &= \quad \frac{M}{|a_n|}\left(\frac{|r|^n-1}{|r| - 1}\right)&&\text{sum the geometric series}\cr\cr \end{alignat} Thus: $$|r^n| \le \frac{M}{|a_n|}\left(\frac{|r|^n-1}{|r| - 1}\right) \tag{3}$$ Multiplying or dividing both sides of an inequality by a positive number does not change the direction of the inequality symbol.
Since, by hypothesis, $\,|r| >1\,$, we have $\,|r| - 1 > 0\,$.
Also, $$|r^n| = |r|^n > 1^n = 1 > 0$$ So, dividing both sides of (3) by $|r^n| = |r|^n$ and multiplying by $\,|r| - 1\,$ gives $$|r| - 1 \le \frac{M}{|a_n|}\frac{|r|^n - 1}{|r|^n} = \frac{M}{|a_n|}\left(1 - \frac{1}{|r|^n}\right)$$ Note:
$$|r| > 1 \quad\Rightarrow\quad |r|^n > 1 \quad\Rightarrow\quad 0 < \frac{1}{|r|^n} < 1 \quad\Rightarrow\quad 0 > -\frac{1}{|r|^n} > -1 \quad\overbrace{\ \ \Rightarrow\ \ }^{\text{add 1}}\quad 1 > 1-\frac{1}{|r|^n} > 0 \quad\Rightarrow\quad 0 < 1-\frac{1}{|r|^n} < 1$$ so $$|r| - 1 \le \frac{M}{|a_n|}$$ from which we get the desired result that $$|r| \le 1 + \frac{M}{|a_n|}$$ QED

Master the ideas from this section