Back in February, I factored the sum and difference of nth odd powers using De Moivre’s Theorem in what is, to my knowledge, a novel approach. To access that proof, click HERE. Additionally, I referenced a single application of my theorem related to evaluating a certain trigonometric series.
In this article, I’d like to extend my theorem to the sum and difference of even powers and explore some more complex applications that would be very difficult to figure out without my theorem.
Let's Dive In!
Unlike in the February proof for the sum/difference of odd powers, in which I was able to find the factorization for the difference \(x^n - a^n\) and then substitute \(-a\) for \(a\) (because \(n\) is presumed odd) to get the sum \(x^n + a^n\), I will have to split the sum/difference of even powers into two cases.
This is the easier case.
Just as in the February proof, using De Moivre’s Theorem, we can prove that:
\[ x^n - a^n = \prod_{k=0}^{n-1} \left( x - a \left( \cos \left( \frac{360k^\circ}{n} \right) + i \sin \left( \frac{360k^\circ}{n} \right) \right) \right) \text{ for any } n \in \mathbb{N} \]
In fact, the only difference between the odd \(n\) cases and the even \(n\) difference case lies in how we group the \( x - a \left( \cos \left( \frac{360k^\circ}{n} \right) + i \sin \left( \frac{360k^\circ}{n} \right) \right) \) terms such that the imaginary component disappears.
We’re looking for pairs \((k_1, k_2)\) selected from the set \(k \in \{ 0, 1, 2, \ldots, n-1 \}\) such that:
\[ \frac{360k_1^\circ}{n} + \frac{360k_2^\circ}{n} = 360^\circ \]
as such would imply that \(-\frac{360k_1^\circ}{n}\) and \(\frac{360k_2^\circ}{n}\) are coterminal angles.
In order for this to be true, \(k_1 + k_2 = n\), meaning we can form pairs \((1, n-1), (2, n-2), (3, n-3), \ldots, \left( \frac{n}{2} - 1, \frac{n}{2} + 1 \right)\) such that the only leftover \(k\) values are \(k = 0\) and \(k = \frac{n}{2}\) since \(n\) is presumed even. Thus, there must be \(\frac{n}{2} - 1\) \((k_1, k_2)\) pairs.
Now we can continue with the rest of the proof from February, finding that the expression
\[ x^2 - 2ax \cos \left( \frac{360k_1^\circ}{n} \right) + a^2 \]
encapsulates both the \(x - a \left( \cos \left( \frac{360k_1^\circ}{n} \right) + i \sin \left( \frac{360k_1^\circ}{n} \right) \right)\) and \(x - a \left( \cos \left( \frac{360k_2^\circ}{n} \right) + i \sin \left( \frac{360k_2^\circ}{n} \right) \right)\) factors for a given pair \((k_1, k_2)\).
Finally, we recognize that the excluded \(k\) values, \(k = 0\) and \(k = \frac{n}{2}\), correspond to the factors \((x - a)\) and \((x + a)\), respectively.
So the final factored form for the difference of even powers is given as:
\[ x^n - a^n = (x - a)(x + a) \prod_{k=1}^{\frac{n}{2} - 1} \left( x^2 - 2ax \cos \left( \frac{360k^\circ}{n} \right) + a^2 \right) \]
This is the harder one, as you’ll notice that unlike the February proof, this one cannot be determined by a simple substitution. Likewise, it's easy to see that there will be no real solutions to \(x^n + a^n = 0\) where \(n\) is even, so this factorization will not conform to the exact structure of the other cases.
We start by considering:
\[ x^n + a^n = 0 \implies x = \left( -a^n \right)^{\frac{1}{n}} = a \left( \cos \left(\frac{180^\circ + 360k^\circ}{n} \right) + i \sin \left( \frac{180^\circ + 360k^\circ}{n} \right) \right) \text{ for } k \in \{ 0, 1, 2, \ldots, n-1 \} \]
So, the factored form is:
\[ x^n + a^n = \prod_{k=0}^{n-1} \left( x - a \left( \cos \left( \frac{180^\circ + 360k^\circ}{n} \right) + i \sin \left( \frac{180^\circ + 360k^\circ}{n} \right) \right) \right) \text{ for any even }n\]
To simplify this further, we use the same strategy of trying to find \((k_1, k_2)\) pairs that result in arguments that are coterminal when the opposite of one is taken. To find these pairs, we have:
\[ \frac{180^\circ + 360k_1^\circ}{n} + \frac{180^\circ + 360k_2^\circ}{n} = 360^\circ \]
or \(k_1 + k_2 = n - 1\). The pairs that satisfy this condition are \((0, n-1), (1, n-2), (2, n-3), \ldots, \left( \frac{n}{2} - 1, \frac{n}{2} \right)\). Just as expected, there are no excluded \(k\) values corresponding to factors that yield real solutions.
Following the same logic as in the even \(n\) difference case and the odd \(n\) cases,
\[ x^n + a^n = \prod_{k=0}^{\frac{n}{2} - 1} \left( x^2 - 2ax \cos \left( \frac{180^\circ + 360k^\circ}{n} \right) + a^2 \right) \text{ for any even } n\]
In the February article, I left as an exercise for the reader to prove that:
\(\sum_{k=1}^{n} \cos\left(\frac{360n^\circ}{2n+1}\right) = -\frac{1}{2}\) for any \(n \in \mathbb{N}\)
Essentially the correct approach was to use the other polynomial factorization for the sum/difference of odd powers (you could’ve used either to arrive at this answer), to solve for the given summation based on the constraint on coefficients. With the addition of the even sum/difference factorizations, one can similarly prove that
\[ \sum_{k=1}^{n} \cos \left( \frac{360k^\circ}{2n+2} \right) = 0 \text{ for any } n \in \mathbb{N} \]
using the other sum/difference polynomial factorization for even powers.
Also interesting is that if you expand out more terms in my theorem’s factorization and equate more coefficients, you can solve for more complicated trigonometric summations. Here are two examples that I’ll again leave as exercises for the reader:
\[ \sum_{1 \leq j < k \leq n} \cos \left( \frac{360j^\circ}{2n+1} \right) \cos \left( \frac{360k^\circ}{2n+1} \right) = \frac{1 - n}{4} \text{ for any } n \in \mathbb{N}\]
And,
\[ \sum_{1 \leq i < j < k \leq n} \cos \left( \frac{360i^\circ}{2n+1} \right) \cos \left( \frac{360j^\circ}{2n+1} \right) \cos \left( \frac{360k^\circ}{2n+1} \right) = \frac{n - 5}{16} \text{ for any } n \in \mathbb{N} \]
Note that these were both obtained using the odd \(n\) power factorizations in the February article. What’s so interesting about these discoveries is that both summations can be expressed as linear functions, one with positive slope, and one with negative slope, curiously. I imagine that expanding out more terms and examining similar summations using the newfound sum/difference of even powers factorizations may yield more of these functions. Perhaps they also follow some sort of rule as one goes down in power coefficients, though, this is just speculation.
Unfortunately, I’m not sure if I’ll be able to return to these discoveries for a little bit, as my summer internship is beginning very shortly and I’m swamped with college application business. Certainly, though, I will get to it as soon as I can!