Factors of Difference of Two Even Powers
Jump to navigation
Jump to search
Theorem
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
- $\ds x^{2 n} - y^{2 n} = \paren {x - y} \paren {x + y} \prod_{k \mathop = 1}^{n - 1} \paren {x^2 - 2 x y \cos \dfrac {k \pi} n + y^2}$
Proof
From Factorisation of $z^n - a$:
- $\ds z^{2 n} - y^{2 n} = \prod_{k \mathop = 0}^{2 n - 1} \paren {x - \alpha^k y}$
where $\alpha$ is a primitive complex $2 n$th roots of unity, for example:
\(\ds \alpha\) | \(=\) | \(\ds e^{2 i \pi / \paren {2 n} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \cos \dfrac {2 \pi} {2 n} + i \sin \dfrac {2 \pi} {2 n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \cos \dfrac \pi n + i \sin \dfrac \pi n\) |
From Complex Roots of Unity occur in Conjugate Pairs:
- $U_{2 n} = \set {1, \tuple {\alpha, \alpha^{2 n - 1} }, \tuple {\alpha^2, \alpha^{2 n - 2} }, \ldots, \tuple {\alpha^k, \alpha^{2 n - k} }, \ldots, \tuple {\alpha^{n - 1}, \alpha^{n + 1} }, -1}$
where $U_{2 n}$ denotes the complex $2 n$th roots of unity:
- $U_{2 n} = \set {z \in \C: z^{2 n} = 1}$
The case $k = 0$ is taken care of by setting $\alpha^0 = 1$, from whence we have the factor $x - y$.
The case $k = n$ is taken care of by setting $\alpha^k = -1$, from whence we have the factor $x + y$.
Taking the product of each of the remaining factors of $x^{2 n} - y^{2 n}$ in pairs:
\(\ds \paren {x - \alpha^k y} \paren {x - \alpha^{2 n - k} y}\) | \(=\) | \(\ds \paren {x - \alpha^k y} \paren {x - \overline {\alpha^k} y}\) | Complex Roots of Unity occur in Conjugate Pairs | |||||||||||
\(\ds \) | \(=\) | \(\ds x^2 - x y \paren {\alpha^k + \overline {\alpha^k} } + \alpha^k y \overline {\alpha^k} y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x^2 - x y \paren {\alpha^k + \overline {\alpha^k} } + \cmod {\alpha^k}^2 y^2\) | Modulus in Terms of Conjugate | |||||||||||
\(\ds \) | \(=\) | \(\ds x^2 - x y \paren {\alpha^k + \overline {\alpha^k} } + y^2\) | Modulus of Complex Root of Unity equals 1 | |||||||||||
\(\ds \) | \(=\) | \(\ds x^2 - x y \paren {\cos \dfrac {k \pi} n + i \sin \dfrac {k \pi} n + \cos \dfrac {k \pi} n - i \sin \dfrac {k \pi} n} + y^2\) | Definition of $\alpha$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x^2 - 2 x y \cos \dfrac {k \pi} n + y^2\) | simplification |
Hence the result.
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 2$: Special Products and Factors: $2.22$