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Polynomial Equations#

Definition: Real Polynomial Equation

A real polynomial equation is a polynomial equation \(P = 0\) where \(P\) is a real polynomial.

Properties#

Theorem: Maximum Number of Roots

The maximum number of distinct roots which the real polynomial equation

\[ A(x) = 0 \]

can have is equal to the degree of \(A\).

Proof

TODO

Theorem: Laguerre's Bound

If the real polynomial \(\sum_{k = 0}^n a_k x^k\) with \(n \ge 2\) can be linearly factorized, then all roots of the equation

\[ \sum_{k = 0}^n a_k x^k = 0 \]

lie within the interval with endpoints

\[ -\frac{a_{n-1}}{n} \pm \frac{n-1}{n}\sqrt{a_{n-1}^2 - \frac{2n}{n-1}a_{n-2}} \]
Proof

TODO

Theorem: Roots and Divisibility

A number \(p \in \mathbb{R}\) is a root of the real polynomial equation

\[ A(x) = 0 \]

if and only if \(A(x)\) is divisible by \((x - p)\).

Proof

TODO

Theorem: Real Polynomial Equations with Integer Coefficients I

If the coefficients \(a_0, \cdots, a_n\) of the real polynomial equation

\[ a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0 \]

are [[TODO|integers]] and it has a [[TODO|rational]] root \(x^\ast = \frac{p}{q}\) (where \(p\) and \(q\) are [[TODO|coprime]]), then \(p\) is a [[TODO|divisor]] of \(a_0\) and \(q\) is a divisor of \(a_n\).

Proof

Proof that \(q\) is a divisor of \(a_n\):

If \(x^\ast = \frac{p}{q}\) is a root of the polynomial equation, then

\[ a_n \frac{p^n}{q^n} + a_{n-1} \frac{p^{n-1}}{q^{n-1}} + \cdots + a_1 \frac{p}{q} + a_0 = 0 \]

Multiply by \(q^{n-1}\).

\[ a_n \frac{p^n}{q} + a_{n-1} p^{n-1} + \cdots + a_1 p q^{n-2} + a_0 q^{n-1} = 0 \]
\[ a_n \frac{p^n}{q} = - a_{n-1} p^{n-1} - \cdots - a_1 p q^{n-2} - a_0 q^{n-1} \]

Since \(p,q\) and \(a_k\) are all integers, the right-hand side must be an integer as well. This means that \(a_n \frac{p^n}{q}\) is an integer, but \(p^n\) and \(q\) have no common divisors, since they are coprime. This means that \(q\) must divide \(a_n\).

Proof that \(p\) is a divisor of \(a_0\):

If \(x^\ast = \frac{p}{q}\) is a root of the polynomial equation, then

\[ a_n \frac{p^n}{q^n} + a_{n-1} \frac{p^{n-1}}{q^{n-1}} + \cdots + a_1 \frac{p}{q} + a_0 = 0 \]

Multiply by \(q^n\).

\[ a_n p^n + a_{n-1} p^{n-1} q + \cdots + a_1 p q^{n-1} + a_0 q^n = 0 \]

Divide by \(p\).

\[ a_n p^n + a_{n-1} p^{n-1} q + \cdots + a_1 p q^{n-1} + a_0 q^n = 0 \]
\[ a_n p^{n-1} + a_{n-1} p^{n-2} q + \cdots + a_1 q^{n-1} + a_0 \frac{q^n}{p} = 0 \]
\[ a_0 \frac{q^n}{p} = - a_n p^{n-1} - a_{n-1} p^{n-2} q - \cdots - a_1 q^{n-1} \]

Once again, \(p,q\) and \(a_k\) are all integers and so the right-hand side must be an integer. This means that \(a_0 \frac{q^n}{p}\) is an integer. The numbers \(q^n\) and \(p\) have no common divisors, since \(q\) and \(p\) are coprime. This means that \(p\) must be a divisor of \(a_0\).

Theorem: Real Polynomial Equations with Integer Coefficients II

If the coefficients of the real polynomial equation

\[ A(x) = 0 \]

are [[TODO|integers]] and it has a [[TODO|rational]] root \(x^\ast = \frac{p}{q}\) (where \(p\) and \(q\) are [[TODO|coprime]]), then for every [[TODO|integer]] \(m\), the number \(A(m)\) is divisible by \(p - mq\).

Proof

TODO

Reciprocal Polynomial Equations#

Definition: Reciprocal Polynomial Equations

A reciprocal polynomial equation is a polynomial equation which can be written either as

\[ a_0 x^{2n} + a_1x^{2n - 1} + \cdots + a_{n-1} x^{n+1} + a_n x^n + a_{n-1} p x^{n-1} + \cdots + a_1 p^{n-1}x + a_0 p^n = 0 \]

or as

\[ a_0 x^{2n + 1} + a_1 x^{2n} + \cdots + a_n x^{n+1} + a_n p x^n + a_{n-1} p^3 x^{n-1} + \cdots + a_1 p^{2n - 1}x + a_0 p^{2n+1} = 0. \]

Definition: Reciprocal Polynomial Equations

Properties#

Algorithm: Degree Reduction for Reciprocal Polynomial Equations of Even Degree

We are given a reciprocal polynomial equation

\[ a_0 x^{2n} + a_1x^{2n - 1} + \cdots + a_{n-1} x^{n+1} + a_n x^n + a_{n-1} p x^{n-1} + \cdots + a_1 p^{n-1}x + a_0 p^n = 0 \]

of degree \(2n\). We can reduce it to a real polynomial equation of degree \(n\) in the following way:

  1. Divide by \(x^n\).

  2. Group the terms appropriately.

\[ a_0\left(x^n + \frac{p^n}{x^n}\right) + a_1 \left(x^{n-1} + \frac{p^{n-1}}{x^{n-1}}\right) + \cdots + a_{n-1} \left( x + \frac{p}{x}\right) + a_n = 0 \]
  1. Substitute \(y = x + \frac{p}{x}\).
\[ a_0 y^n + a_1 y^{n-1} + \cdots + a_{n-1} y + a_n = 0 \]

Theorem: Roots of Reciprocal Polynomial Equations of Odd Degree

Every reciprocal polynomial equation

\[ a_0 x^{2n + 1} + a_1 x^{2n} + \cdots + a_n x^{n+1} + a_n p x^n + a_{n-1} p^3 x^{n-1} + \cdots + a_1 p^{2n - 1}x + a_0 p^{2n+1} = 0. \]

of odd degree \(2n+1\) has the root \(x = -p\).

Proof

TODO

Theorem: Reduction of Reciprocal Polynomial Equations of Odd Degree

Every reciprocal polynomial equation

\[ a_0 x^{2n + 1} + a_1 x^{2n} + \cdots + a_n x^{n+1} + a_n p x^n + a_{n-1} p^3 x^{n-1} + \cdots + a_1 p^{2n - 1}x + a_0 p^{2n+1} = 0. \]

of odd degree \(2n+1\) can be reduced to a reciprocal polynomial equation of even degree \(2n\) by dividing it by \(x+p\).

Proof

TODO