Generalized evaluation of ordered sums of cosine products in an arithmetic sequence

1. Introduction

The purpose of this paper is to derive a general formula for ordered sums of cosine products where the arguments progress in an arithmetic sequence. The paper begins by exploring an alternate form of the difference of odd powers, using this as a stepping stone to establish a connection with trigonometric identities. From there, the focus shifts to deriving the generalized formula for ordered sums of cosine products, supported by recursive relations and inductive proofs.

The structure of the paper is as follows:

Difference of Odd Powers Alternate Form: Introduces an alternative factorization of the difference of odd powers, setting up the trigonometric framework used later.
Equating Alternate Difference of Odd Powers Formula with Common One: Proves the equivalence between the classical polynomial identity and the alternate form through term expansion and regrouping, then derives generalized coefficients for even and odd powered terms.
**Odd \(r\)**: Establishes a recursive relationship for the derived coefficients of ODD-powered terms and validates a combinatorial closed-form formula through inductive reasoning.
**Even \(r\)**: Establishes a recursive relationship for the derived coefficients of EVEN-powered terms and validates a combinatorial closed-form formula through inductive reasoning.
Conclusion: Summarizes the key results and implies avenues for future research and application.

2. Difference of Odd Powers Alternate Form

We begin with the polynomial \(y = x^{2q+1} - a^{2q+1}\), for \(q \in \mathbb{N}_0\), and \(a \in \mathbb{R}\). In order to factor such a polynomial, we will first find its zeros:

\[0 = x^{2q+1} - a^{2q+1} \implies x^{2q+1} = a^{2q+1} \implies x = \bigl(a^{2q+1}\bigr)^{\tfrac{1}{2q+1}}\]

Using exponent rules to simplify \((a^{2q+1})^{1/(2q+1)}\) would eliminate any complex solutions to the equation. Therefore, we will instead use De Moivre's theorem to allow values of \(x\) (including complex ones!) such that \(y = 0 = x^{2q+1} - a^{2q+1}\).

Since \(a \in \mathbb{R}\), we have:

\[x = \bigl(a^{2q+1}\bigr)^{\tfrac{1}{2q+1}} = |a^{2q+1}|^{\tfrac{1}{2q+1}}\!\left(\cos\!\tfrac{360t^{\circ}}{2q+1} + i\sin\!\tfrac{360t^{\circ}}{2q+1}\right) = a\!\left(\cos\!\tfrac{360t^{\circ}}{2q+1} + i\sin\!\tfrac{360t^{\circ}}{2q+1}\right)\]

for \(t \in \{0, 1, 2, \ldots, 2q\}\), and arguments in degree measure.

Now, to obtain the factors of \(y\), we have:

\[y = \prod_{t=0}^{2q}\!\left(x - a\!\left(\cos\!\tfrac{360t^{\circ}}{2q+1} + i\sin\!\tfrac{360t^{\circ}}{2q+1}\right)\right)\]

But this can be simplified further. The argument in each factor, \(\tfrac{360t^{\circ}}{2q+1}\), will yield angles \(0^{\circ}, \tfrac{360(1)^{\circ}}{2q+1}, \tfrac{360(2)^{\circ}}{2q+1}, \ldots, \tfrac{360(2q-1)^{\circ}}{2q+1}, \tfrac{360(2q)^{\circ}}{2q+1}\). Note that, since \(2q+1\) is odd, one can form \(\tfrac{2q}{2} = q\) distinct pairs of angles that sum to \(360^{\circ}\) utilizing values of \(t\) that sum to \(2q+1\).

For instance, for \(t=1\), \(t=2q\), \(\tfrac{360(1)^{\circ}}{2q+1} + \tfrac{360(2q)^{\circ}}{2q+1} = 360^{\circ}\). Since pairs of angles sum to \(360^{\circ}\), for a given pair \((t_1, t_2)\), where \(t_1 + t_2 = 2q + 1\):

\[\tfrac{360t_1^{\circ}}{2q+1} + \tfrac{360t_2^{\circ}}{2q+1} = 360^{\circ} \implies \tfrac{360t_2^{\circ}}{2q+1} = 360^{\circ} - \tfrac{360t_1^{\circ}}{2q+1}\]

Therefore, \(-\tfrac{360t_1^{\circ}}{2q+1}\) and \(\tfrac{360t_2^{\circ}}{2q+1}\) are coterminal angles. By extent, for the same pair \((t_1, t_2)\), the factor \(x - a(\cos\tfrac{360t_1^{\circ}}{2q+1} + i\sin\tfrac{360t_1^{\circ}}{2q+1})\) must accompany another factor,

\[x - a\!\left(\cos\tfrac{360t_2^{\circ}}{2q+1} + i\sin\tfrac{360t_2^{\circ}}{2q+1}\right) = x - a\!\left(\cos\tfrac{360t_1^{\circ}}{2q+1} - i\sin\tfrac{360t_1^{\circ}}{2q+1}\right)\]

Now, we can group the factored expressions for \((t_1, t_2)\) together using a difference of squares:

\[\left(x - a\cos\tfrac{360t_1^{\circ}}{2q+1}\right)^{2} - \left(ia\sin\tfrac{360t_1^{\circ}}{2q+1}\right)^{2}\]

Expanding, and using the identity \(\cos^{2}(u) + \sin^{2}(u) = 1\), we have \(x^{2} - 2ax\cos\tfrac{360t_1^{\circ}}{2q+1} + a^{2}\). For every given pair \((t_1, t_2)\) in the factored form of \(y\), of which we know there are \(q\), we can express the product of the \(t_1\) and \(t_2\) terms as \(x^{2} - 2ax\cos\tfrac{360t_1^{\circ}}{2q+1} + a^{2}\). This leads to the final factored form of \(y\):

\[\boxed{\ y = x^{2q+1} - a^{2q+1} = (x - a)\prod_{t=1}^{q}\!\left(x^{2} - 2ax\cos\tfrac{360t^{\circ}}{2q+1} + a^{2}\right)\ }\]

3. Equating the Alternate Formula with the Common One

Replacing variables, we have the identity

\[a^{2q+1} - b^{2q+1} = (a - b)\prod_{t=1}^{q}\!\left(a^{2} - 2ab\cos\tfrac{360t^{\circ}}{2q+1} + b^{2}\right),\ \forall q \in \mathbb{N}.\]

There exists an equivalent form for this formula:

\[a^{2q+1} - b^{2q+1} = (a - b)\!\left(a^{2q} + a^{2q-1}b + a^{2q-2}b^{2} + \cdots + a^{2}b^{2q-2} + ab^{2q-1} + b^{2q}\right)\ \forall q \in \mathbb{N}.\]

So we have the equivalence

\[\prod_{t=1}^{q}\!\left(a^{2} - 2ab\cos\tfrac{360t^{\circ}}{2q+1} + b^{2}\right) = a^{2q} + a^{2q-1}b + a^{2q-2}b^{2} + \cdots + ab^{2q-1} + b^{2q}.\]

We can substitute \(c_t = \cos\tfrac{360t^{\circ}}{2q+1}\) in the LHS yielding

\[\prod_{t=1}^{q}\!\bigl((a^{2} + b^{2}) - 2ab\,c_t\bigr) = a^{2q} + a^{2q-1}b + \cdots + b^{2q}.\]

Expanding the LHS as a sum over subsets and writing

\[\sum_{1 \le t_1 < t_2 < \cdots < t_r \le q} c_{t_1}c_{t_2}\cdots c_{t_r} = C_{q,r},\ \forall q, r \in \mathbb{N},\]

the LHS becomes

\[\sum_{r=0}^{q}(a^{2}+b^{2})^{q-r}\,(-2ab)^{r}\,C_{q,r}.\]

Expanding the \((a^{2}+b^{2})^{q-r}\) component via the Binomial theorem and incorporating the \((ab)^{r}\) multiplier, we get a sum of \(a^{2q-r}b^{r}\) terms. Regrouping by the power of \(b\), the \((r+1)\)th term in the expansion is, for odd \(r\):

\[-\Bigl[\,2\tbinom{q-1}{(r-1)/2}C_{q,1} + 8\tbinom{q-3}{(r-1)/2 - 1}C_{q,3} + 32\tbinom{q-5}{(r-1)/2 - 2}C_{q,5} + \cdots + 2^{r-2}\tbinom{q-(r-2)}{1}C_{q,r-2} + 2^{r}C_{q,r}\,\Bigr]\,a^{2q-r}b^{r}\]

and for even \(r\):

\[\Bigl[\,\tbinom{q}{r/2} + 4\tbinom{q-2}{r/2 - 1}C_{q,2} + 16\tbinom{q-4}{r/2 - 2}C_{q,4} + \cdots + 2^{r-2}\tbinom{q-(r-2)}{1}C_{q,r-2} + 2^{r}C_{q,r}\,\Bigr]\,a^{2q-r}b^{r}\]

where the top argument of the binomials iterates \(-2\) across each term in the coefficients and the bottom argument iterates \(-1\) across each term. Since both expressions equal \(a^{2q-r}b^{r}\) with coefficient \(1\) (by the equivalence we began with), we can develop recursive formulae for \(C_{q,r}\).

4. Odd \(r\)

All further calculations assume \(r \in \{2i + 1 \mid i \in \mathbb{N}\}\). We first aim to develop a recursive formula for \(C_{q,r}\) based on previous coefficients:

\[1 = -\!\left[2\tbinom{q-1}{(r-1)/2}C_{q,1} + 8\tbinom{q-3}{(r-1)/2 - 1}C_{q,3} + \cdots + 2^{r}C_{q,r}\right]\]

\[C_{q,r} = \frac{-\!\left[\,\displaystyle\sum_{m=0}^{(r-3)/2} 2^{2m+1} C_{q,2m+1}\tbinom{q-1-2m}{(r-1)/2 - m}\,\right] + 1}{2^{r}}\]

We now assume an equivalent closed-form expression for \(C_{q,r}\) (discovered initially through bulk computation),

\[C_{q,r} = \frac{(-1)^{(r+1)/2}}{2^{r}}\binom{q - 1 - (r-1)/2}{(r-1)/2},\]

which holds for all odd \(r \le k\). By induction, if we show that this formula holds for \(k+2\), we'll have proven general equivalence. Note that \(C_{q,1} = -\tfrac{1}{2}\) for both formulae. This is our base case.

First, we rewrite the proposed equation for \(C_{q,r}\) with \(r = 2m+1\):

\[C_{q,2m+1} = \frac{(-1)^{m+1}}{2^{2m+1}}\binom{q - 1 - m}{m}.\]

Substituting into the recursive formula for \(r = k+2\) and simplifying yields:

\[C_{q,k+2} = \frac{-\tfrac{1}{((k+1)/2)!}\!\left[\,\displaystyle\sum_{m=0}^{(k-1)/2}(-1)^{m+1}\tbinom{(k+1)/2}{m}\!\prod_{n=1}^{(k+1)/2}(q - n - m)\,\right] + ((k+1)/2)!}{2^{k+2}}\]

We now simplify the sum in the numerator. The approach relies on shifting the \(m\)-offset within the intra-sum \((q - n - m)\) products to a common \((q - n - (k+1)/2)\) so the product can be factored out of the sum. That requires a correction term \(P_m\). We separate the \(q\)-independent constant from the shift and call the strictly \(q\)-dependent part \(Q_m\):

\[\sum_{m=0}^{(k-1)/2}(-1)^{m+1}\tbinom{(k+1)/2}{m}\prod_{n=1}^{(k+1)/2}(q - n - m) = \left[\prod_{n=1}^{(k+1)/2}\!\left(q - n - \tfrac{k+1}{2}\right)\right]\!\left[\sum_{m=0}^{(k-1)/2}(-1)^{m+1}\tbinom{(k+1)/2}{m}\right] + P_m\]

\(P_m\) must cancel the shift to a uniform offset of \((k+1)/2\). Therefore,

\[P_m = \sum_{m=1}^{(k-1)/2}\!\left[\sum_{1 \le n_1 < \cdots < n_m \le (k+1)/2}\prod_{j=1}^{m}\!\left(q - n_j - \tfrac{k+1}{2}\right)\right]\!\left[\sum_{n=0}^{(k-1)/2}\!\left(\tfrac{k+1}{2} - n\right)^{(k+1)/2 - m}\!(-1)^{n+1}\tbinom{(k+1)/2}{n}\right] + \sum_{m=0}^{(k-1)/2}\!\left(\tfrac{k+1}{2} - m\right)^{(k+1)/2}\!(-1)^{m+1}\tbinom{(k+1)/2}{m}\]

Note that both the second internal summation of \(Q_m\) and the constant extracted from \(P_m\) are very similar to the formula for Stirling Numbers of the second kind, \(S(a, b)\), which denotes the number of ways to partition a set of \(a\) objects into \(b\) non-empty subsets:

\[S(a, b) = \frac{1}{b!}\sum_{i=0}^{b}(-1)^{b-i}\tbinom{b}{i}\,i^{a} = \begin{cases} 0 & \text{if } a < b, \\ 1 & \text{if } a = b. \end{cases}\]

Re-indexing the second internal summation of \(Q_m\) in the RHS with \(n = (k+1)/2 - i\), we get \(-((k+1)/2)!\,S((k+1)/2 - m,\,(k+1)/2)\). Since \(m > 0\) and therefore \((k+1)/2 - m < (k+1)/2\) for all terms in \(P_m\), the inner sum \(S(\cdot, \cdot) = 0\), collapsing the whole contribution of \(P_m\)'s first component to \(0\).

Note too that the constant term extracted from \(P_m\) is simply \(-((k+1)/2)!\,S((k+1)/2 - m,\,(k+1)/2)\) at \(m = 0\), so

\[\sum_{m=0}^{(k-1)/2}\!\left(\tfrac{k+1}{2} - m\right)^{(k+1)/2}\!(-1)^{m+1}\tbinom{(k+1)/2}{m} = -\!\left(\tfrac{k+1}{2}\right)!\,S\!\left(\tfrac{k+1}{2}, \tfrac{k+1}{2}\right) = -\!\left(\tfrac{k+1}{2}\right)!.\]

Simplifying the first term is relatively straightforward. By the Binomial Expansion formula,

\[\sum_{m=0}^{(k-1)/2}(-1)^{m+1}\tbinom{(k+1)/2}{m} = -(1 - 1)^{(k+1)/2} + (-1)^{(k+1)/2} = (-1)^{(k+1)/2}.\]

Thus, the first term becomes \((-1)^{(k+1)/2}\prod_{n=1}^{(k+1)/2}\!\left(q - n - \tfrac{k+1}{2}\right)\). Plugging back into \(C_{q,k+2}\):

\[C_{q,k+2} = \frac{-\tfrac{1}{((k+1)/2)!}\!\left[(-1)^{(k+1)/2}\!\prod_{n=1}^{(k+1)/2}\!\left(q - n - \tfrac{k+1}{2}\right) - \left(\tfrac{k+1}{2}\right)!\right] + \left(\tfrac{k+1}{2}\right)!}{2^{k+2}} = \frac{(-1)^{(k+3)/2}}{2^{k+2}}\binom{q - 1 - (k+1)/2}{(k+1)/2},\]

thus showing that our proposed formula \(C_{q,r}\) holds for \(r = k+2\) and justifying our inductive reasoning. To summarize, for \(r \in \{2i+1 \mid i \in \mathbb{N}\}\), \(q \in \mathbb{N}\):

\[\boxed{\ C_{q,r} = \sum_{1 \le t_1 < t_2 < \cdots < t_r \le q}\prod_{j=1}^{r}\cos\!\left(\tfrac{360t_j^{\circ}}{2q+1}\right) = \frac{(-1)^{(r+1)/2}}{2^{r}}\binom{q - 1 - (r-1)/2}{(r-1)/2}\ }\]

An intriguing formula comparing ordered sums of cosine products to alternating binomial coefficients decreasing exponentially in value. Now let's do it for the even \(r\) case.

5. Even \(r\)

This case follows nearly the same logic. All further calculations assume \(r \in \{2i \mid i \in \mathbb{N}_{0}\}\). We first aim to develop a recursive formula for \(C_{q,r}\) based on previous coefficients:

\[1 = \tbinom{q}{r/2} + 4\tbinom{q-2}{r/2 - 1}C_{q,2} + 16\tbinom{q-4}{r/2 - 2}C_{q,4} + \cdots + 2^{r-2}\tbinom{q-(r-2)}{1}C_{q,r-2} + 2^{r}C_{q,r}\]

\[C_{q,r} = \frac{-\displaystyle\sum_{m=0}^{(r-2)/2} 2^{2m} C_{q,2m}\tbinom{q-2m}{r/2 - m} + 1}{2^{r}}\]

We now assume an equivalent closed-form expression (also discovered through bulk computation):

\[C_{q,r} = \frac{(-1)^{r/2}}{2^{r}}\binom{q - r/2}{r/2},\]

which holds for all even \(r \le k\). By induction, if we show that this formula holds for \(k+2\), we'll have proven general equivalence. Note \(C_{q,0} = 1\) for both formulae. This is our base case.

Substituting \(C_{q,2m} = \tfrac{(-1)^{m}}{2^{2m}}\binom{q-m}{m}\) into the recursive formula for \(r = k+2\) and simplifying yields, by the identical Stirling-number argument and Binomial expansion:

\[C_{q,k+2} = \frac{-\tfrac{1}{((k+2)/2)!}\!\left[(-1)^{k/2}\!\prod_{n=0}^{k/2}\!\left(q - n - \tfrac{k+2}{2}\right) + \left(\tfrac{k+2}{2}\right)!\right] - \left(\tfrac{k+2}{2}\right)!}{2^{k+2}} = \frac{(-1)^{(k+2)/2}}{2^{k+2}}\binom{q - (k+2)/2}{(k+2)/2},\]

Thus showing that our proposed formula \(C_{q,r}\) holds for \(r = k+2\) and justifying our inductive reasoning. To summarize, for \(r \in \{2i \mid i \in \mathbb{N}_0\}\), \(q \in \mathbb{N}\):

\[\boxed{\ C_{q,r} = \sum_{1 \le t_1 < \cdots < t_r \le q}\prod_{j=1}^{r}\cos\!\left(\tfrac{360t_j^{\circ}}{2q+1}\right) = \frac{(-1)^{r/2}}{2^{r}}\binom{q - r/2}{r/2}\ }\]

6. Conclusion

This paper has successfully derived a closed-form general formula for the ordered sums of cosine products where the arguments progress in an arithmetic sequence. These results connect the sums of cosine products to alternating binomial coefficients, which decay exponentially, thus highlighting a combinatorial aspect within trigonometric identities. The study also uncovered connections to Stirling numbers of the second kind, further emphasizing the deep combinatorial structure underlying these sums. This research not only simplifies the evaluation of such sums but also lays the groundwork for future explorations into related mathematical fields — potential extensions could involve exploring other trigonometric identities or applying these findings to problems in combinatorics and analysis.

Worked examples recovered from the general formula: \(C_{q,1} = -\tfrac{1}{2}\) (Article 1 conjecture); \(C_{q,2} = \tfrac{1-q}{4}\) and \(C_{q,3} = \tfrac{q-5}{16}\) (Article 5 exercises).