Question

Prove that \(x^{2 m+1}-a^{2 m+1}\), where \(m\) is an integer \(\geq 1\), can be written as $$ x^{2 m+1}-a^{2 m+1}=(x-a) \prod_{r=1}^{m}\left[x^{2}-2 a x \cos \left(\frac{2 \pi r}{2 m+1}\right)+a^{2}\right] $$

Short answer

First, use roots of unity to factorize. Then group conjugate pairs and apply trigonometric identities to factorize completely.

Step-by-step solution

- Start with the Polynomial Identity

Consider the given polynomial expression, \ \(P(x) = x^{2 m+1} - a^{2 m+1}\).\ Notice that this can be factored by employing roots of unity.

- Use Roots of Unity

Recall that we can express \(P(x)\) in terms of roots of unity. Let \( \omega = e^{2 \pi i / (2m+1)} \) be a primitive \((2m+1)^{th}\) root of unity. The roots of \ \(a^{2m+1} \) are given by \ \(a\omega^k\) for \ \(k = 0, 1, 2, ..., 2m\).

- Decompose the Polynomial

Given that \ \(P(a\omega^k) = 0 \) for \ \(k = 0, 1, 2, ..., 2m\), the polynomial can be written as: \ \[ P(x) = (x - a)(x - a\omega)(x - a\omega^2)...(x - a\omega^{2m}) \]

- Simplify for Conjugate Pairs

Pair the conjugates in the product form to simplify it: \ \[ (x - a)(x - a\omega)(x - a\omega^2)...(x - a\omega^{2m}) = (x - a) \prod_{k=1}^{m} (x - a\omega^k)(x - a\omega^{-k}) \] where \ \( \omega^{-k} = \omega^{2m+1-k} \).

- Rewrite Using Cosine Terms

Recognize that \ \( \omega^k = cos(2 \pi k/(2m+1)) + i sin(2 \pi k/(2m+1)) \), which allows us to utilize the sum of squares for conjugates \ \[( x - a \omega^k )( x - a \omega^{-k} ) = x^2 - a x(\omega^k + \omega^{-k}) + a^2 \].\ Note that \ \( \omega^k + \omega^{-k} = 2 cos(2 \pi k / (2m+1)) \). So, the expression simplifies to: \ \[( x^2 - 2ax cos(2 \pi k / (2m+1)) + a^2) \].

- Combine Results

Combining all the steps, we get: \ \[ x^{2m+1} - a^{2m+1} = ( x - a ) \prod_{k=1}^{m} ( x^2 - 2ax cos(2 \pi k / (2m+1)) + a^2) \] which matches the given form.

Key concepts

Roots of Unity

Roots of unity are special complex numbers that represent the solutions to the equation \(z^n = 1\). These roots play a critical role in various areas of mathematics, including polynomial factorization and Fourier transforms.
To understand roots of unity, let's consider the equation \(z^{2m+1} = 1\). The solutions to this equation are called the \(2m+1\)-th roots of unity.
A fundamental root of unity is often denoted as \(\omega = e^{2 \pi i / (2m+1)}\). All other roots can be represented as powers of \(\omega\), specifically \(\omega^k\) where \(k = 0, 1, 2, ..., 2m\). For instance, \(\omega, \omega^2, \omega^3,...,\omega^{2m}\) are all roots of unity, and they evenly distribute on the complex unit circle.

Cosine Terms

Cosine terms in polynomial factorization arise naturally when working with roots of unity. From Euler's formula, we have \(\omega^k = \cos(\frac{2 \pi k}{2m+1}) + i \sin(\frac{2 \pi k}{2m+1})\).
When dealing with conjugate pairs in the polynomial, the imaginary components cancel out. This simplifies the problem significantly and brings in cosine terms:
\[ \omega^k + \omega^{-k} = \cos(\frac{2 \pi k}{2m+1}) + i \sin(\frac{2 \pi k}{2m+1}) + \cos(\frac{2 \pi k}{2m+1}) - i \sin(\frac{2 \pi k}{2m+1}) = 2 \cos(\frac{2 \pi k}{2m+1}) \]
Converting these forms helps in representing the factorized polynomial in terms of cosine, reducing the complexity of the algebraic manipulation.

Complex Numbers

Complex numbers extend the real numbers to solve equations that do not have real solutions. They have the form \(a + bi\), where \(i\) is the imaginary unit satisfying \(i^2 = -1\).
Polynomials often yield complex roots, especially when they are expressed in terms of roots of unity. In our polynomial factorization problem, we use the roots of unity, represented as complex numbers of the form \(\omega = e^{2 \pi i / (2m+1)}\), helping us to break down and factorize the polynomial.
These complex representations allow the use of simplification techniques, such as pairing conjugate terms, to manage and reduce the complexity of polynomial transformations and expressions.

Primitive Roots

Primitive roots are particular solutions among the roots of unity that can generate all other roots through exponentiation. Specifically, for roots of unity, a primitive \(n\)-th root of unity \(\omega\) satisfies \(\omega^k\) for \(k = 0, 1, ..., n-1\), yielding all \(n\) distinct roots.
In our factorization problem, \(\omega = e^{2 \pi i / (2m+1)}\) serves as a primitive \( (2m+1)\)-th root of unity. This means that \(\omega, \omega^2, \omega^3,...,\omega^{2m}\) are distinct and cover all the roots for the equation \(z^{2m+1} = 1\).
Understanding primitive roots is crucial for rearranging and simplifying polynomials involving high degrees, as they provide the structural backbone for decomposing complex polynomials into achievable factors.