On the Generalized Evaluation of Ordered Sums of Cosine Products with
Arguments Progressing in an Arithmetic Sequence
Abraham Lobsenz
October 12, 2024
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:
1. Difference of Odd Powers Alternate Form: Introduces an alternative factorization of the difference of odd powers,
setting up the trigonometric framework used later.
2. 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.
3. Odd r: Establishes a recursive relationship for the derived coefficients of ODD-powered terms and validates a combinatorial
closed-form formula through inductive reasoning.
4. Even r: Establishes a recursive relationship for the derived coefficients of EVEN-powered terms and validates a combina-
torial closed-form formula through inductive reasoning.
5. 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 ∈ N
0
, and a ∈ R. In order to factor such a polynomial, we will first find
its zeros:
0 = x
2q+1
− a
2q+1
=⇒ x
2q+1
= a
2q+1
=⇒ x =
a
2q+1
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 ∈ R, we have:
x =
a
2q+1
1
2q+1
= |a
2q+1
|
1
2q+1
cos
360t
◦
2q + 1
+ i sin
360t
◦
2q + 1
= a
cos
360t
◦
2q + 1
+ i sin
360t
◦
2q + 1
,
for t = {0, 1, 2, . . . , 2q}, and arguments in degree measure.
Now, to obtain the factors of y, we have:
1
y =
2q
Y
t=0
x − a
cos
360t
◦
2q + 1
+ i sin
360t
◦
2q + 1
But this can be simplified further. The argument in each factor,
360t
◦
2q+1
, will yield angles 0
◦
,
360(1)
◦
2q+1
,
360(2)
◦
2q+1
, . . . ,
360(2q−1)
◦
2q+1
,
360(2q)
◦
2q+1
.
Note that, since 2q + 1 is odd, one can form
2q
2
= q distinct pairs of angles that sum to 360
◦
utilizing values of t that sum to
2q + 1.
For instance, for t = 1, t = 2q,
360(1)
◦
2q+1
+
360(2q)
◦
2q+1
= 360
◦
. Since pairs of angles sum to 360
◦
, for a given pair (t
1
, t
2
), where
t
1
+ t
2
= 2q + 1,
360t
◦
1
2q + 1
+
360t
◦
2
2q + 1
= 360
◦
=⇒
360t
◦
2
2q + 1
= 360
◦
−
360t
◦
1
2q + 1
Therefore, −
360t
◦
1
2q+1
and
360t
◦
2
2q+1
are coterminal angles.
By extent, for the same pair (t
1
, t
2
), the factor x − a
cos
360t
◦
1
2q+1
+ i sin
360t
◦
1
2q+1
must accompany another factor,
x−a
cos
360t
◦
2
2q + 1
+ i sin
360t
◦
2
2q + 1
= x−a
cos
−
360t
◦
1
2q + 1
+ i sin
−
360t
◦
1
2q + 1
= x−a
cos
360t
◦
1
2q + 1
− i sin
360t
◦
1
2q + 1
Now, we can group the factored expressions for (t
1
, t
2
) together using a difference of squares:
x − a
cos
360t
◦
1
2q + 1
+ i sin
360t
◦
1
2q + 1
x − a
cos
360t
◦
1
2q + 1
− i sin
360t
◦
1
2q + 1
=
x − a cos
360t
◦
1
2q + 1
2
−
ia sin
360t
◦
1
2q + 1
2
Expanding, and using the identity cos
2
(u) + sin
2
(u) = 1, we have:
x
2
− 2ax cos
360t
◦
1
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
360t
◦
1
2q + 1
+ a
2
This leads to the final factored form of y:
y = x
2q+1
− a
2q+1
= (x − a)
q
Y
t=1
x
2
− 2ax cos
360t
◦
2q + 1
+ a
2
3 Equating Alternate Difference of Odd Powers Formula with Common One
Replacing variables, we have the identity,
a
2q+1
− b
2q+1
= (a − b)
q
Y
t=1
a
2
− 2ab cos
360t
◦
2q + 1
+ b
2
∀q ∈ N. There exists an equivalent form for this formula
1
:
2
a
2q+1
− b
2q+1
= (a − b)
a
2q
+ a
2q−1
b + a
2q− 2
b
2
+ . . . + a
2
b
2q− 2
+ ab
2q− 1
+ b
2q
∀q ∈ N. So, we have the equivalence:
q
Y
t=1
a
2
− 2ab cos
360t
◦
2q + 1
+ b
2
=
a
2q
+ a
2q− 1
b + a
2q−2
b
2
+ . . . + a
2
b
2q−2
+ ab
2q− 1
+ b
2q
We can substitute c
t
= cos
360t
◦
2q+1
in the LHS yielding,
q
Y
t=1
((a
2
+ b
2
) − 2abc
t
) =
a
2q
+ a
2q−1
b + a
2q− 2
b
2
+ . . . + a
2
b
2q− 2
+ ab
2q− 1
+ b
2q
Now we can focus on each term in the expanded LHS, yielding the summed terms,
(a
2
+ b
2
)
q
+
+(a
2
+ b
2
)
q −1
· −2ab
q
X
t
1
=1
c
t
1
+(a
2
+ b
2
)
q −2
· 4a
2
b
2
X
1≤t
1
<t
2
≤q
c
t
1
c
t
2
+(a
2
+ b
2
)
q−3
· −8a
3
b
3
X
1≤t
1
<t
2
<t
3
≤q
c
t
1
c
t
2
c
t
3
.
.
.
+(a
2
+ b
2
)
0
· (−2ab)
q
c
1
c
2
c
3
. . . c
q
For convenience, we’ll state that,
X
1≤t
1
<t
2
<t
3
<...<t
r
≤q
c
t
1
c
t
2
c
t
3
. . . c
t
r
= C
q ,r
∀q, r ∈ N. Expanding the exponentiated (a
2
+ b
2
) components and incorporating the (ab)
q
multiplier present within all terms
expect the first yields,
(a
2q
+
q
1
a
2q− 2
b
2
+
q
2
a
2q− 4
b
4
+ . . . +
q
2
a
4
b
2q− 4
+
q
1
a
2
b
2q− 2
+ b
2q
)
−2(a
2q− 1
b +
q − 1
1
a
n−4
b
3
+
q − 1
2
a
2q− 5
b
5
+ . . . +
q − 1
2
a
5
b
2q− 5
+
q − 1
1
a
3
b
2q−3
+ ab
2q− 1
)C
q ,1
+4(a
2q− 2
b
2
+
q − 2
1
a
2q−4
b
4
+
q − 2
2
a
2q− 6
b
6
+ . . . +
q − 2
2
a
6
b
2q− 6
+
q − 2
1
a
4
b
2q−4
+ a
2
b
2q− 2
)C
q ,2
−8(a
2q− 3
b
3
+
q − 3
1
a
2q−5
b
5
+
q − 3
2
a
2q− 7
b
7
+ . . . +
q − 3
2
a
7
b
2q− 7
+
q − 3
1
a
5
b
2q−5
+ a
3
b
2q− 3
)C
q ,3
3
.
.
.
+(−2)
q
(ab)
q
C
q ,q
Note that binomial symmetry properties are used for clarity. Now, we recognize that many of the same a
2q−r
b
r
terms present
in the RHS are beginning to emerge. The strategy from here is clear: Combine terms in the expanded LHS to find a
2q−r
b
r
coefficients, which, per the RHS, should all be equal to 1. We regroup to find the following expression:
a
2q
− 2C
q ,1
a
2q− 1
b +
q
1
+ 4C
q ,2
a
2q− 2
b
2
−
2
q − 1
1
C
q ,1
+ 8C
q ,3
a
2q− 3
b
3
−
q
2
+ 4
q − 2
1
C
q ,2
+ 16C
q ,4
a
2q− 4
b
4
. . .
+
q
2
+ 4
q − 2
1
C
q ,2
+ 16C
q ,4
b
2q− 4
a
4
−
2
q − 1
1
C
q ,2
+ 8C
q ,3
b
2q− 3
a
3
+
q
1
+ 4C
q ,2
b
2q−2
a
2
− (2C
q ,1
) b
2q− 1
a + b
2q
The (r + 1)th term can be generalized as:
−
2
q − 1
r−1
2
C
q ,1
+ 8
q − 3
r−1
2
− 1
C
q ,3
+ 32
q − 5
r−1
2
− 2
C
q ,5
. . . + 2
e−2
q − (r − 2)
1
C
q ,r−2
+ 2
r
C
q ,r
a
2q−r
b
r
, r ∈ {2i+1 | i ∈ N}
And,
q
r
2
+ 4
q − 2
r
2
− 1
C
q ,2
+ 16
q − 4
r
2
− 2
C
q ,4
+ . . . + 2
r−2
q − (r − 2)
1
C
q ,r−2
+ 2
r
C
q ,r
a
2q− r
b
r
, r ∈ {2i | i ∈ N
0
}
Where the top argument of the binomials iterates −2 across each term in the coefficients and the bottom argument iterates −1
across each term in the coefficients. Note that both of these expressions are equal to a
2q− r
b
r
due to the equivalence we began
with, so their coefficients are equal to 1. This fact can be used to develop recursive formulae for C
q ,r
sums.
4 Odd r
All further calculations assume r ∈ {2i + 1 | i ∈ N}. We first aim to develop a recursive formula for C
q ,r
based on previous
coefficients.
1 = −
2
q − 1
r−1
2
C
q ,1
+ 8
q − 3
r−1
2
− 1
C
q ,3
+ . . . + 2
r
C
q ,r
C
q ,r
=
−
2
q −1
r−1
2
C
q ,1
+ 8
q−3
r−1
2
−1
C
q ,3
+ . . . + 2
r−2
q −(r−2)
1
C
q ,r−2
+ 1
2
r
C
q ,r
=
−
P
r−3
2
m=0
2
2m+1
C
2m+1
q−1−2m
r−1
2
−m
+ 1
2
e
We now assume an equivalent closed-form expression for C
q ,r
(discovered initially through bulk computation),
C
q ,r
=
(−1)
r+1
2
2
r
q − 1 −
r−1
2
r−1
2
which holds for all odd r ≤ k. By induction, if we show that this formula holds for k + 2, we’ll have proven general equivalence.
Note that C
q ,1
= −
1
2
for both formulae. This is our base case.
4
First, we rewrite the proposed equation for C
q ,r
with r = 2m + 1,
C
2m+1
=
(−1)
m+1
2
2m+1
q − 1 − m
m
Now we substitute C
2m+1
into the recursive formula for r = k + 2,
C
q ,k+2
=
−
P
k−1
2
m=0
2
2m+1
C
2m+1
q −1−2m
k+1
2
−m
+ 1
2
k+2
=
−
P
k−1
2
m=0
2
2m+1
(−1)
m+1
2
2m+1
q −1−m
m
q−1−2m
k+1
2
−m
+ 1
2
k+2
And simplify,
C
q ,k+2
=
−
P
k−1
2
m=0
2
2m+1
(−1)
m+1
2
2m+1
·
(q −1−m)!
m!(q −1−2m)!
·
(q −1−2m)!
(
k+1
2
−m
)
!
(
q −1−
k+1
2
−m
)
!
+ 1
2
k+2
C
q ,k+2
=
−
P
k−1
2
m=0
(−1)
m+1
(q −1−m)!(q−1−2m)!
m!(q −1−2m)!
(
k+1
2
−m
)
!
(
q −1−
k+1
2
−m
)
!
+ 1
2
k+2
C
q ,k+2
=
−
P
k−1
2
m=0
1
m!
(
k+1
2
−m
)
!
·
(
k+1
2
)
!
(
k+1
2
)
!
·
(−1)
m+1
(q −1−m)!(q−1−2m)!
(q −1−2m)!
(
q −1−
k+1
2
−m
)
!
+ 1
2
k+2
C
q ,k+2
=
−
1
(
k+1
2
)
!
P
k−1
2
m=0
(−1)
m+1
(
k+1
2
)
!
m!
(
k+1
2
−m
)
!
·
(q −1−m)!
(
q−1−
k+1
2
−m
)
!
+
k+1
2
!
2
k+2
C
q ,k+2
=
−
1
(
k+1
2
)
!
P
k−1
2
m=0
(−1)
m+1
k+1
2
m
Q
k+1
2
n=1
(q − n − m) +
k+1
2
!
2
k+2
Now we focus on simplifying the sum in the numerator. The following approach relies on shifting the m-offset within the intra-
sum (q − n − m) products to a common (q − n −
k+1
2
) such that the product can be factored out of the sum. In order to do
this, we need a correction term P
m
to account for the shift. Finally, its helpful to separate the q-independent constant from this
correction term for later consideration. We’ll call the new strictly q−dependent part of the correction term Q
m
. We progress
like so,
k−1
2
X
m=0
(−1)
m+1
k+1
2
m
k+1
2
Y
n=1
(q − n − m) =
k−1
2
X
m=0
(−1)
m+1
k+1
2
m
k+1
2
Y
n=1
(q − n −
k + 1
2
) + P
m
=
k+1
2
Y
n=1
q − n −
k + 1
2
k−1
2
X
m=0
(−1)
m+1
k+1
2
m
+ P
m
P
m
needs to cancel out the shift to a uniform offset of
k+1
2
. Therefore,
P
m
=
k−1
2
X
m=0
X
1≤n
1
<n
2
<n
3
<...<n
m
≤
k+1
2
m
Y
j=1
q − n
j
−
k + 1
2
k−1
2
X
n=0
k + 1
2
− n
k+1
2
−m
(−1)
n+1
k+1
2
n
5
=
k−1
2
X
m=1
X
1≤n
1
<n
2
<n
3
<...<n
m
≤
k+1
2
m
Y
j=1
q − n
j
−
k + 1
2
k−1
2
X
n=0
k + 1
2
− n
k+1
2
−m
(−1)
n+1
k+1
2
n
+
k−1
2
X
m=0
k + 1
2
− m
k+1
2
(−1)
m+1
k+1
2
m
Therefore, our re-written sum is expressed as:
k−1
2
X
m=0
(−1)
m+1
k+1
2
m
k+1
2
Y
n=1
(q − n − m) =
k+1
2
Y
n=1
q − n −
k + 1
2
k−1
2
X
m=0
(−1)
m+1
k+1
2
m
+
k−1
2
X
m=1
X
1≤n
1
<n
2
<n
3
<...<n
m
≤
k+1
2
m
Y
j=1
q − n
j
−
k + 1
2
k−1
2
X
n=0
k + 1
2
− n
k+1
2
−m
(−1)
n+1
k+1
2
n
+
k−1
2
X
m=0
k + 1
2
− m
k+1
2
(−1)
m+1
k+1
2
m
Note that both the second internal summation of Q
m
and the constant we 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
2
. The formula for S(a, b) is,
S(a, b) =
1
b!
b
X
i=0
(−1)
b−i
b
i
i
a
=
(
0 if a < b
1 if a = b
Re-indexing the second internal summation of Q
m
in the RHS with n =
k+1
2
− i, we have,
k−1
2
X
n=0
k + 1
2
− n
k+1
2
−m
(−1)
n+1
k+1
2
n
=
k+1
2
X
i=1
i
k+1
2
−m
(−1)
(
k+1
2
−i+1
)
k+1
2
k+1
2
− i
= −
k+1
2
X
i=1
(i)
k+1
2
−m
(−1)
(
k+1
2
−i
)
k+1
2
i
=
−
k+1
2
X
i=0
i
k+1
2
−m
(−1)
(
k+1
2
−i
)
k+1
2
i
+ 0 = −
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
of the RHS,
k−1
2
X
m=1
X
1≤n
1
<n
2
<n
3
<...<i
m
≤
k+1
2
m
Y
j=1
q − n
j
−
k + 1
2
k−1
2
X
n=0
k + 1
2
− n
m
(−1)
n+1
k+1
2
n
=
k−1
2
X
m=1
X
1≤n
1
<n
2
<n
3
<...<i
m
≤
k+1
2
m
Y
j=1
q − n
j
−
k + 1
2
−
k + 1
2
!S
m,
k + 1
2
6
=
k−1
2
X
m=1
X
1≤n
1
<n
2
<n
3
<...<i
m
≤
k+1
2
m
Y
j=1
q − n
j
−
k + 1
2
· 0 = 0
Note that the constant term extracted from the P
m
is simply −
k+1
2
!S
k+1
2
− m,
k+1
2
for m = 0, so,
k−1
2
X
m=0
k + 1
2
− m
k+1
2
−0
(−1)
m+1
k+1
2
m
= −
k + 1
2
!S
k + 1
2
,
k + 1
2
= −
k + 1
2
!
Simplifying the first term is relatively straight forward. Note that,
k−1
2
X
m=0
(−1)
m+1
k+1
2
m
= −
k+1
2
X
m=0
(−1)
m
(1)
k+1
2
−m
k+1
2
m
− (−1)
k+3
2
= −(1 − 1)
k+1
2
+ (−1)
k+1
2
= (−1)
k+1
2
By the Binomial Expansion formula
3
. Thus, the first term becomes,
k+1
2
Y
n=1
q − n −
k + 1
2
k−1
2
X
m=0
(−1)
m+1
k+1
2
m
= (−1)
k+1
2
k+1
2
Y
n=1
q − n −
k + 1
2
So, we’ve shown that,
k−1
2
X
m=0
(−1)
m+1
k+1
2
m
k+1
2
Y
n=1
(q − n − m) =
k+1
2
Y
n=1
q − n −
k + 1
2
k−1
2
X
m=0
(−1)
m+1
k+1
2
m
+
k−1
2
X
m=1
X
1≤n
1
<n
2
<n
3
<...<n
m
≤
k+1
2
m
Y
j=1
q − n
j
−
k + 1
2
k−1
2
X
n=0
k + 1
2
− n
k+1
2
−m
(−1)
n+1
k+1
2
n
+
k−1
2
X
m=0
k + 1
2
− m
k+1
2
(−1)
m+1
k+1
2
m
= (−1)
k+1
2
k+1
2
Y
n=1
q − n −
k + 1
2
−
k + 1
2
!
Plugging this expression back into the numerator we have,
C
q ,k+2
=
−
1
(
k+1
2
)
!
P
k−1
2
m=0
(−1)
m+1
k+1
2
m
Q
k+1
2
n=1
(q − n − m) +
k+1
2
!
2
k+2
=
−
1
(
k+1
2
)
!
(−1)
k+1
2
Q
k+1
2
n=1
q − n −
k+1
2
−
k+1
2
! +
k+1
2
!
2
k+2
=
7
−(−1)
k+1
2
·
(
q−1−
k+1
2
)
!
(
k+1
2
)
!(q −1−(k+1))!
2
k+2
=
(−1)
k+3
2
2
k+2
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, then, we’ve found that for r ∈ {2i + 1 | i ∈ N}, q ∈ N,
C
q ,r
=
X
1≤t
1
<t
2
<t
3
<...<t
r
≤q
cos
360t
◦
1
2q + 1
cos
360t
◦
2
2q + 1
cos
360t
◦
3
2q + 1
. . . cos
360t
◦
r
2q + 1
=
(−1)
r+1
2
2
r
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 as outlined in the odd r case. All further calculations assume r ∈ {2i | i ∈ N
0
}. We first
aim to develop a recursive formula for C
q ,r
based on previous coefficients.
1 =
q
r
2
+ 4
q − 2
r
2
− 1
C
q ,2
+ 16
q − 4
r
2
− 2
C
q ,4
+ . . . + 2
r−2
q − (r − 2)
1
C
q ,r−2
+ 2
r
C
q ,r
C
q ,r
=
−
q
r
2
+ 4
q −2
r
2
−1
C
q ,2
+ 16
q−4
r
2
−2
C
q ,4
+ . . . + 2
r−2
q −(r−2)
1
C
q ,r−2
+ 1
2
r
C
q ,r
=
−
P
r−2
2
m=0
2
2m
C
2m
q −2m
r
2
−m
+ 1
2
r
We now assume an equivalent closed-form expression for C
q ,r
(also discovered initially through bulk computation),
C
q ,r
=
(−1)
r
2
2
r
q −
r
2
r
2
which holds for all even r ≤ k. By induction, if we show that this formula holds for k + 2, we’ll have proven general equivalence.
Note that C
q ,0
= 1 for both formulae. This is our base case.
First, we rewrite the proposed equation for C
q ,r
with r = 2m,
C
2m
=
(−1)
m
2
2m
q − m
m
Now we substitute this expression for C
2m
into the recursive formula for r = k + 2,
C
q ,k+2
=
−
P
k
2
m=0
2
2m
C
2m
q −2m
k+2
2
−m
+ 1
2
k+2
8
=
−
P
k
2
m=0
2
2m
(−1)
m
2
2m
q−m
m
q −2m
k+2
2
−m
+ 1
2
k+2
And simplify,
C
q ,k+2
=
−
P
k
2
m=0
2
2m
(−1)
m
2
2m
·
(q −m)!
m!(q −2m)!
·
(q −2m)!
(
k+2
2
−m
)
!
(
q−
k+2
2
−m
)
!
+ 1
2
k+2
C
q ,k+2
=
−
P
k
2
m=0
(−1)
m
(q −m)!(q−2m)!
m!(q −2m)!
(
k+2
2
−m
)
!
(
q −
k+2
2
−m
)
!
+ 1
2
k+2
C
q ,k+2
=
−
P
k
2
m=0
1
m!
(
k+2
2
−m
)
!
·
(
k+2
2
)
!
(
k+2
2
)
!
·
(−1)
m
(q −m)!(q−2m)!
(q −2m)!
(
q −
k+1
2
−m
)
!
+ 1
2
k+2
C
q ,k+2
=
−
1
(
k+2
2
)
!
P
k
2
m=0
(−1)
m
(
k+2
2
)
!
m!
(
k+2
2
−m
)
!
·
(q −m)!
(
q −
k+2
2
−m
)
!
−
k+2
2
!
2
k+2
C
q ,k+2
=
−
1
(
k+2
2
)
!
P
k−1
2
m=0
(−1)
m
k+2
2
m
Q
k
2
n=0
(q − n − m) −
k+2
2
!
2
k+2
Now we focus on simplifying the sum in the numerator. The following approach relies on shifting the m-offset within the intra-
sum (q − n − m) products to a common (q − n −
k+2
2
) such that the product can be factored out of the sum. In order to do
this, we need a correction term P
m
to account for the shift. Finally, its helpful to separate the q-independent constant from this
correction term for later consideration. We’ll call the new strictly q−dependent part of the correction term Q
m
. We progress
like so,
k
2
X
m=0
(−1)
m
k+2
2
m
k
2
Y
n=0
(q − n − m) =
k
2
X
m=0
(−1)
m
k+2
2
m
k
2
Y
n=0
(q − n −
k + 2
2
) + P
m
=
k
2
Y
n=0
q − n −
k + 2
2
k
2
X
m=0
(−1)
m
k+2
2
m
+ P
m
P
m
needs to cancel out the shift to a uniform offset of
k+2
2
. Therefore,
P
m
=
k
2
X
m=0
X
1≤n
1
<n
2
<n
3
<...<n
m
≤
k
2
m
Y
j=1
q − n
j
−
k + 2
2
k
2
X
n=0
k + 2
2
− n
k+2
2
−m
(−1)
n
k+2
2
n
=
k
2
X
m=1
X
1≤n
1
<n
2
<n
3
<...<n
m
≤
k
2
m
Y
j=1
q − n
j
−
k + 2
2
k
2
X
n=0
k + 2
2
− n
k+2
2
−m
(−1)
n
k+2
2
n
+
k
2
X
m=0
k + 2
2
− m
k+2
2
(−1)
m
k+2
2
m
Therefore, our re-written sum is expressed as:
9
k
2
X
m=0
(−1)
m
k+2
2
m
k
2
Y
n=0
(q − n − m) =
k
2
Y
n=0
q − n −
k + 2
2
k
2
X
m=0
(−1)
m
k+2
2
m
+
k
2
X
m=1
X
1≤n
1
<n
2
<n
3
<...<n
m
≤
k
2
m
Y
j=1
q − n
j
−
k + 2
2
k
2
X
n=0
k + 2
2
− n
k+2
2
−m
(−1)
n
k+2
2
n
+
k
2
X
m=0
k + 2
2
− m
k+2
2
(−1)
m
k+2
2
m
Note that both the second internal summation of Q
m
and the constant we extracted from P
m
are again very similar to the
formula for S(a, b) given in the odd r case. Re-indexing the second internal summation of Q
m
in the RHS with n =
k+2
2
− i, we
have,
k
2
X
n=0
k + 2
2
− n
k+2
2
−m
(−1)
n
k+2
2
n
=
k+2
2
X
i=1
i
k+2
2
−m
(−1)
(
k+2
2
−i
)
k+2
2
k+2
2
− i
=
k+2
2
X
i=1
i
k+2
2
−m
(−1)
(
k+2
2
−i
)
k+2
2
i
=
k+2
2
X
i=0
i
k+2
2
−m
(−1)
(
k+2
2
−i
)
k+2
2
i
− 0 =
k + 2
2
!S
k + 2
2
− m,
k + 2
2
Since m > 0 and therefore
k+2
2
− m <
k+2
2
for all terms in P
m
of the RHS,
k
2
X
m=1
X
1≤n
1
<n
2
<n
3
<...<n
m
≤
k
2
m
Y
j=1
q − n
j
−
k + 2
2
k
2
X
n=0
k + 2
2
− n
k+2
2
−m
(−1)
n
k+2
2
n
=
k
2
X
m=1
X
1≤n
1
<n
2
<n
3
<...<n
m
≤
k
2
m
Y
j=1
q − n
j
−
k + 2
2
k + 2
2
!S
k + 2
2
− m,
k + 2
2
=
k
2
X
m=1
X
1≤n
1
<n
2
<n
3
<...<n
m
≤
k
2
m
Y
j=1
q − n
j
−
k + 2
2
· 0 = 0
Note that the constant term extracted from the P
m
is simply
k+2
2
!S
k+2
2
− m,
k+2
2
for m = 0, so,
k
2
X
m=0
k + 2
2
− m
k+2
2
−0
(−1)
m
k+2
2
m
=
k + 2
2
!S
k + 2
2
,
k + 2
2
=
k + 2
2
!
Simplifying the first term is relatively straight forward. Note that,
k
2
X
m=0
(−1)
m
k+2
2
m
=
k+2
2
X
m=0
(−1)
m
(1)
k+2
2
−m
k+2
2
m
− (−1)
k+2
2
= (1 − 1)
k+2
2
+ (−1)
k
2
= (−1)
k
2
By the Binomial Expansion formula
3
. Thus, the first term becomes,
10
k
2
Y
n=0
q − n −
k + 2
2
k
2
X
m=0
(−1)
m
k+2
2
m
= (−1)
k+2
2
k
2
Y
n=0
q − n −
k + 2
2
So, we’ve shown that,
k
2
X
m=0
(−1)
m
k+2
2
m
k
2
Y
n=0
(q − n − m) =
k
2
Y
n=0
q − n −
k + 2
2
k
2
X
m=0
(−1)
m
k+2
2
m
+
k
2
X
m=1
X
1≤n
1
<n
2
<n
3
<...<n
m
≤
k
2
m
Y
j=1
q − n
j
−
k + 2
2
k
2
X
n=0
k + 2
2
− n
k+2
2
−m
(−1)
n
k+2
2
n
+
k
2
X
m=0
k + 2
2
− m
k+2
2
(−1)
m
k+2
2
m
= (−1)
k
2
k
2
Y
n=0
q − n −
k + 2
2
+
k + 2
2
!
Plugging this expression back into the numerator we have,
C
q ,k+2
=
−
1
(
k+2
2
)
!
P
k−1
2
m=0
(−1)
m
k+2
2
m
Q
k
2
n=0
(q − n − m) −
k+2
2
!
2
k+2
=
=
−
1
(
k+2
2
)
!
(−1)
k
2
Q
k
2
n=0
q − n −
k+2
2
+
k+2
2
! −
k+2
2
!
2
k+2
−(−1)
k
2
·
(
q−
k+2
2
)
!
(
k+2
2
)
!(q −(k+2))!
2
k+2
=
(−1)
k+2
2
2
k+2
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, then, we’ve found that for r ∈ {2i | i ∈ N
0
}, q ∈ N,
C
q ,r
=
X
1≤t
1
<t
2
<t
3
<...<t
r
≤q
cos
360t
◦
1
2q + 1
cos
360t
◦
2
2q + 1
cos
360t
◦
3
2q + 1
. . . cos
360t
◦
r
2q + 1
=
(−1)
r
2
2
r
q −
r
2
r
2
11
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. The derived formulas are:
C
q ,r
=
X
1≤t
1
<t
2
<t
3
<...<t
r
≤q
cos
360t
◦
1
2q + 1
cos
360t
◦
2
2q + 1
cos
360t
◦
3
2q + 1
. . . cos
360t
◦
r
2q + 1
=
(−1)
r+1
2
2
r
q − 1 −
r−1
2
r−1
2
for odd r
And,
C
q ,r
=
X
1≤t
1
<t
2
<t
3
<...<t
r
≤q
cos
360t
◦
1
2q + 1
cos
360t
◦
2
2q + 1
cos
360t
◦
3
2q + 1
. . . cos
360t
◦
r
2q + 1
=
(−1)
r
2
2
r
q −
r
2
r
2
for even r
These results connect the sums of cosine products to alternating binomial coefficients, which decay exponentially, thus high-
lighting 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.
7 References:
1
Art of Problem Solving. (n.d.). Sum and difference of powers. Retrieved August 27, 2024.
2
Wikipedia. (n.d.). Stirling numbers of the second kind. In Wikipedia, The Free Encyclopedia. Retrieved August 27, 2024.
3
Wikipedia. (n.d.). Binomial theorem. In Wikipedia, The Free Encyclopedia. Retrieved August 27, 2024.
12
