Basic trigonometric power sums with applications

Abstract

We present the transformation of several sums of positive integer powers of the sine and cosine into non-trigonometric combinatorial forms. The results are applied to the derivation of generating functions and to the number of the closed walks on a path and in a cycle.

Appendices

Appendix 1

In the introduction it was stated that the result of Berndt and Yeap’s for the sum over even powers of cotangent, viz. (1.2), was imprecise and that a better formulation was given by (1.3). Here, we prove this by referring to [3]. To enable the reader to develop an understanding of how polynomials in k arise when evaluating finite sums of powers of the cotangent, we shall also show how the formula is implemented for specific values of n, which is also lacking in [3].

Broadly speaking, Berndt and Yeap derive (1.3) via the third case considered in [3, Thm. 2.1]. The other two cases will be discussed in a future work. The theorem deals with the contour integration of the function

$$\begin{aligned} f(z)=\cot ^m(\pi z) \cot (\pi (hz-a))\cot (\pi (kz-b)) \end{aligned}$$

over a positively oriented rectangle with vertices at $\pm i R$ and $1 \pm iR$, where $R>\epsilon$, and possessing semi-circular indentations at 0 and 1 of radius $\epsilon$, where $\epsilon < \min \{ (h-1+a)/h,(k-1+b)/k \}$. The third case is represented by $a=b=0$ and hence f(z) becomes

$$\begin{aligned} f(z) = (hk)^{-1} (\pi z)^{-m-2} \left( \sum _{j=0}^{\infty } a(j) x^j \right) ^m \sum _{\mu =0}^{\infty } a(\mu ) (h' x)^{\mu } \sum _{\nu =0}^{\infty } a(\nu ) (k' x)^{\nu }, \end{aligned}$$

(8.1)

where $a_j = (-1)^j 2^{2j} B_{2j}/(2j)!$, $x = (\pi z)^2$, $h' = h^2$ and $k' = k^2$. From (8.1) we see that there is a pole of order $m+2$ at $z = 0$. Therefore, the aim is to evaluate the residue of f(z) at $z = 0$, which is given by

$$\begin{aligned} \mathrm{Res} \,f(z) \left| _{z=0}= \frac{1}{(m+1)!} \, \frac{d^{m+1}}{dz^{m+1}}\, \left( z^{m+2} f(z) \right) \right| _{z=0}. \end{aligned}$$

(8.2)

Before we can evaluate this, we need to evaluate the product of the infinite series on the rhs of (8.1). Berndt and Yeap proceed by introducing coefficients $C(j_1,\ldots ,j_m,\mu ,\nu )$, which are not given explicitly, but that they arise when a sum over all $(m+2)$-tuples $(j_1,\ldots , j_m,\mu ,\nu )$ is evaluated under the condition that $2\left( \sum _{i=1}^m j_i +\mu + \right.$ $\left. \nu \right) = m+1$. This, however, leads to the imprecision in (1.3). Here we adopt a different approach based on extending the Cauchy product formula [32].

We begin by multiplying one of the series in the power by the penultimate series in (8.1). If we denote this product as $P_1$, then by the Cauchy product formula, it can be expressed as

$$\begin{aligned} P_1= \sum _{j=0}^{\infty } x^j A_1(j), \end{aligned}$$

where $A_1(j) = \sum _{\ell _1=0}^{j} a(\ell _1) h^{'\,\ell _1} a(j-\ell _1)$. Now we multiply $P_1$ by the final series in (8.1), which yields

$$\begin{aligned} P_2= P_1 \; \sum _{\nu =0}^{\infty } a(\nu ) k^{'\,\nu } x^{\nu }= \sum _{j=0}^{\infty } x^j \sum _{\ell _1=0}^{j}a(\ell _1) k^{'r\, \ell _1} A_1(j-\ell _1) = \sum _{j=0}^{\infty } x^j A_2(j-\ell _1), \end{aligned}$$

and

$$\begin{aligned} A_2(j) = \sum _{\ell _1=0}^{j} \sum _{\ell _2=0}^{j-\ell _1} a(\ell _1) \, a(\ell _2)\, k^{'\,\ell _1} h^{'\,\ell _2} \, a(j-\ell _1-\ell _2). \end{aligned}$$

Next we multiply $P_2$ by another series in the power to obtain $P_3$, obtaining

$$\begin{aligned} P_3= P_2 \sum _{j=0}^{\infty } a(j) x^{j}= \sum _{j=0}^{\infty } x^j \sum _{\ell _1=0}^{j} a(\ell _1) A_2(j-\ell _1)= \sum _{j=0}^{\infty } A_3(j) x^j, \end{aligned}$$

where

$$\begin{aligned} A_3(j) =\sum _{\ell _1=0}^{j}\sum _{\ell _2=0}^{j-\ell _1} \sum _{\ell _3=0}^{j-\ell _1-\ell _2} h^{'\,\ell _2}k^{'\, \ell _3}\, a(\ell _1)\, a(\ell _2)\, a(\ell _3) \, a(j-\ell _1-\ell _2-\ell _3). \end{aligned}$$

Continuing this process until all the series have been multiplied out, we eventually arrive at

$$\begin{aligned} \left( \sum _{j=0}^{\infty } a(j) x^j \right) ^m \sum _{\mu =0}^{\infty } a(\mu ) (h' x)^{\mu } \sum _{\nu =0}^{\infty } a(\nu ) (k' x)^{\nu } = \sum _{j=0}^{\infty } A_{m+1}(j) \,x^j, \end{aligned}$$

where the coefficients are given by

$$\begin{aligned} A_{m+1}(j)= & {} \sum _{\ell _1=0}^{j}\sum _{\ell _2=0}^{j-\ell _1} \sum _{\ell _3=0}^{j-\ell _1-\ell _2} \cdots \sum _{\ell _{m+1}=0}^{j-\ell _1-\ell _2-\cdots -\ell _{m}} h^{'\,\ell _m}k^{'\,\ell _{m+1}} \nonumber \\&\times \, \prod _{i=1}^{m+1} a(\ell _i) \, a(j-\ell _1-\ell _2-\dots -\ell _m-\ell _{m+1}). \end{aligned}$$

(8.3)

If we let $\ell _s=\sum _{i=1}^{m+1} \ell _{i}$ and replace the various terms in (8.3) by their values in f(z), then we find that

$$\begin{aligned} f(z)= & {} (\pi z)^{-m-2}\sum _{j=0}^{\infty } (-1)^j (2\pi z)^{2j} \sum _{\ell _1,\ell _2,\ell _3, \ldots ,\ell _{m+1}=0}^{j,j-\ell _1,j-\ell _1-\ell _2,\ldots ,j-\ell _s+\ell _{m+1}} h^{'\,2\ell _m-1}\,k^{'\,2\ell _{m+1}-1} \nonumber \\&\times \, \prod _{i=1}^{m+1} \frac{B_{2\ell _i}}{(2\ell _i)!} \, \frac{B_{2(j-\ell _s)}}{(2(j-\ell _s))!}. \end{aligned}$$

(8.4)

Hence there is a pole of order $m+2$ at $z=0$. Introducing the above result into (8.2), we see that there is only a residue when $m+1$ is equal to one of the even powers of z inside the summation over j. Therefore, m must be odd for f(z) to yield a residue. Introducing (8.4) into (8.2), with m replaced by $2n-1$, where n is a positive integer, yields

$$\begin{aligned} \mathrm{Res} \,f(z) \Big |_{z=0}= & {} \frac{(-1)^n \,2^{2n} }{\pi } \sum _{\ell _1=0,\ell _2=0, \ldots ,\ell _{2n}=0}^{n, n-\ell _1,\ldots ,n-\ell _s+\ell _{2n}} h^{'\,2\ell _{2n-1}-1} k^{'\,2\ell _{2n}-1}\\&\times \, \prod _{i=1}^{2n} \frac{B_{2\ell _i}}{(2\ell _i)!} \, \frac{B_{2n-2\ell _s}}{(2n-2\ell _s)!}, \end{aligned}$$

where $\ell _s =\sum _{i=1}^{2n} \ell _i$.

To obtain a finite sum over powers of the cotangent, we need to consider the entire contour around f(z). This means that there are simple poles at $z = (j+a)/h$ and $z = (r+b)/k$, where j and r are non-negative integers such that $0<j+a<h$ and $0<r+b<k$. Since this is the third case, where $a=b=0$, these become $z=j/h$ and $z=r/k$. Moreover, because $h=1$, we can disregard the poles at $z=j$ , while r ranges from 1 to $k - 1$. In addition, by noting that $\lim _{y \rightarrow \infty } \cot (c(x\pm iy)+d)= \pm i$, for $c>0$ and d real, Berndt and Yeap are able to evaluate the contour integral directly, thereby obtaining

$$\begin{aligned} \frac{1}{2\pi \, i}\int _C f(z) \, dz= \frac{(-1)^n}{\pi }. \end{aligned}$$

By applying Cauchy’s residue theorem, we finally arrive at (1.3).

To conclude this appendix, let us now discuss the implementation of (1.3) for $n = 2$. Then the $j_i$ range from $j_1$ to $j_4$. For $n-j_s$ to be non-zero, we require some of the $j_i$ to be zero, whereas according to the Berndt–Yeap result given by (1.2), they should be greater than zero. For $n =2$, $j_0$ can be equal to 0, 1, or 2. When $j_0=2$, all the other $j_i$’s must vanish and the sum in (1.3) contributes the value $-(-1)^2 (2^4) k^3 B_4/4!$, which in turn equals $-k^3/45$ since $B_4 = -1/30$. When $j_0=1$, either the remaining $j_i$ equal unity or $2-j_s$ equals unity. Hence there are four possibilities, each yielding the same contribution. The total contribution for $j_0 =1$ becomes $4 k ((-1) (2^2) B_2/2!)^2$ or 4k / 9 since $B_2 = 1/6$. When $j_0=0$, we have two separate cases. In the first of these cases either one of the remaining $j_i$ or $2-j_s$ equals 2. Since there are four possibilities, we obtain a contribution of $4 \cdot 2^4 B_4/(4!\, k)$ for this case. For the second case one of the remaining $j_i$ or $2-j_s$ is equal to unity and another one must also be equal to unity. Since there are effectively four variables including $2-j_s$, this means there are $\left( {\begin{array}{c}4\\ 2\end{array}}\right)$ or 6 combinations. Therefore, the contribution from the second case is $6 (-2^2 B_2/2!)^2/k$. Combining the two cases yields the total contribution for $j_0=0$, which is $(-4/45 +2/3)k^{-1}$. Thus, (1.3) for $n=2$ gives

$$\begin{aligned} \frac{1}{k} \sum _{r=1}^{k-1} \cot ^4 \left( \frac{\pi r}{k} \right) =1- (-k^3/45 +4 k/9 +26/45 k). \end{aligned}$$

After a little algebra, one eventually obtains

$$\begin{aligned} \sum _{r=1}^{k-1} \cot ^4 \left( \frac{\pi r}{k} \right) = \frac{1}{45} \, (k-1) (k-2) (k^2 +3k-13), \end{aligned}$$

which appears as Corollary 2.6a in Berndt and Yeap [3]. In a similar fashion one can calculate the results for $n = 3$ and $n = 4$, the details of which are not presented here. After a little algebra, one finds that

$$\begin{aligned} \sum _{r=1}^{k-1} \cot ^6 \left( \frac{\pi r}{k} \right) = \frac{1}{945} \, (k-1) (k-2) \bigr ( 2k^4+6 k^3-28 k^2 -96 k +251 \bigr ), \end{aligned}$$

(8.5)

and

$$\begin{aligned} \sum _{r=1}^{k-1} \cot ^8 \left( \frac{\pi r}{k} \right)= & {} \frac{1}{14175} \, (k-1) (k-2) \left( 3k^6+9 k^5-59 k^4 \right. \\&- \left. \,195 k^3+ 457 k^2 +1761 k -3551 \right) . \end{aligned}$$

By using a different method, Gessel has obtained (8.5), which, aside from a phase factor, appears as $q_6(n)$ in [16]. It should also be mentioned that beyond $n = 4$, the calculations become cumbersome due to the rapidly increasing number of combinations when the $j_i$ are summed to n. For these values of n, a computer program will be needed to evaluate (1.3).

Appendix 2

In this appendix we consider multiplying and dividing the argument in the trigonometric powers of the sums C(m, n) and S(m, n) by 5 or what is referred to as the $\ell =5$ case according to the terminology of Sect. 4. In so doing, the material presented here should enable the reader to consider other values of $\ell$, although we shall see that higher values of $\ell$ are not as tractable as the cases studied in Sect. 3.

To investigate the $\ell =5$ case, we require the following general identity:

$$\begin{aligned} \sum _{j=1}^{\ell } e^{2 \pi ij k/\ell } = {\left\{ \begin{array}{ll} \ell , &{} \quad k \equiv 0\; (\mathrm{mod}\; \ell ), \\ 0 , &{} \quad \mathrm{otherwise}. \end{array}\right. } \end{aligned}$$

(9.1)

Multiplying and dividing the argument in the cosine power of C(m, n) as defined in Sect. 4 by 5, we obtain

$$\begin{aligned} C(m,n) = \sum _{k=0,5,10,\ldots }^{5n-5} \cos ^{2m} \left( \frac{k \pi }{5 n} \right) . \end{aligned}$$

Next we put $\ell = 5$ in (9.1) and introduce it into the above equation. After a little algebra, we arrive at

$$\begin{aligned} C(m,n)= \frac{1}{5}\sum _{k=0}^{5n-1} \left( 2\, \cos \left( \frac{2 \pi k}{5} \right) + 2 \, \cos \left( \frac{4 \pi k}{5} \right) +1 \right) \cos ^{2m} \left( \frac{k \pi }{5 n} \right) . \end{aligned}$$

(9.2)

The last sum on the rhs of (9.2) is C(m, 5n). Hence we are left with two distinct sums. To isolate these sums, we need to consider an even multiple of 5, e.g., $\ell = 10$, since we observed that the basic trigonometric sums in Sect. 4 turned out to be reducible when $\ell$ was even.

By multiplying and dividing the argument of the cosine power in C(m, n) by 10, we find that

$$\begin{aligned} C(m,n) = \sum _{k=0,10,20,\ldots }^{10n-10} \cos ^{2m} \left( \frac{k \pi }{10 n} \right) . \end{aligned}$$

Now we introduce the $\ell = 10$ version of (9.1) into the above result, which after a little algebra yields

$$\begin{aligned} C(m,n)= & {} \frac{1}{10} \sum _{k=0}^{10n-1} \left( 2\, \cos \left( \frac{\pi k}{5} \right) + 2\, \cos \left( \frac{2 \pi k}{5} \right) + 2\, \cos \left( \frac{3 \pi k}{5} \right) \right. \\&\left. + \, 2 \, \cos \left( \frac{4 \pi k}{5} \right) +1 +(-1)^k \right) \cos ^{2m} \left( \frac{k \pi }{10 n} \right) . \end{aligned}$$

The above result can be simplified by introducing the trigonometric identity for the sum of two cosines, which is given as No. 1.314(3) in [17]. In this instance we sum the first and fourth cosines on the rhs and then the second and third cosines. Then we obtain

$$\begin{aligned} 10\, C(m,n)= & {} \sum _{k=0}^{10n-1} \left( 4\, \cos \left( \frac{\pi k}{2} \right) \cos \left( \frac{3 \pi k}{10} \right) + 4\, \cos \left( \frac{\pi k}{2} \right) \cos \left( \frac{\pi k}{10} \right) \right. \nonumber \\&\left. + \; 1 +(-1)^k \right) \cos ^{2m} \left( \frac{k \pi }{10 n} \right) . \end{aligned}$$

(9.3)

In (9.3) all terms with odd values of k vanish, so we can replace k by 2k, which leads to

$$\begin{aligned} 10\, C(m,n)-2\, C(m,5n) = \sum _{k=0}^{5n-1} (-1)^k \left( 4\, \cos \left( \frac{3 \pi k}{5} \right) + 4\, \cos \left( \frac{\pi k}{5} \right) \right) \cos ^{2m} \left( \frac{k \pi }{5 n} \right) . \end{aligned}$$

Alternatively, the above result can be written as

$$\begin{aligned} 10 \,C(m,n)-2\, C(m,5n) = \sum _{k=0}^{5n-1} \left( 4\, \cos \left( \frac{2 \pi k}{5} \right) + 4\, \cos \left( \frac{4 \pi k}{5} \right) \right) \cos ^{2m} \left( \frac{k \pi }{5 n} \right) . \end{aligned}$$

(9.4)

This, however, is twice (9.2). Therefore, the $\ell = 10$ case reduces to the $\ell = 5$ case, just as we observed in the $\ell = 3$ and $\ell = 6$ cases. Worse still, the two series involving $\cos (2\pi k/5)$ and $\cos (4\pi k /5)$ cannot be decoupled. That is, extra information is required before each of these series can be evaluated separately. However, we can express (9.2) as

$$\begin{aligned} C(m,n)= \frac{1}{5}\sum _{k=0}^{5n-1} \left( 2\, \cos \left( \frac{2 \pi k}{5} \right) + 2 \, \cos \left( \frac{6 \pi k}{5} \right) +1 \right) \cos ^{2m} \left( \frac{k \pi }{5 n} \right) . \end{aligned}$$

(9.5)

By applying the identity for the sum of two cosines, we finally arrive at the following basic cosine power sum:

$$\begin{aligned} \sum _{k=0}^{5n-1} \cos \left( \frac{ 2\pi k}{5} \right) \cos \left( \frac{4 \pi k}{5} \right) \cos ^{2m} \left( \frac{k \pi }{5 n} \right) = \frac{1}{4} \, \left( 5 C(m,n) -C(m,5n) \right) . \end{aligned}$$

(9.6)

In actual fact the above result is not very surprising because the product of the cosines external to the cosine power is given by

$$\begin{aligned} \cos \left( \frac{ 2\pi k}{5} \right) \cos \left( \frac{4 \pi k}{5} \right) = {\left\{ \begin{array}{ll} 1, &{} k \equiv 0,\; (\mathrm{mod} \; 5), \\ -1/4 , &{} \mathrm{otherwise}. \end{array}\right. } \end{aligned}$$

It is also interesting to note that we cannot use the alternating version of C(m, n) to decouple the sums in (9.5). From (3.3) we have

$$\begin{aligned} \sum _{k=0}^{n-1} (-1)^k \cos ^{2m} \left( \frac{5k \pi }{5n} \right)= & {} \sum _{k=0,5,\ldots }^{5n-5} \cos \left( \frac{ \pi k}{5}\right) \cos ^{2m} \left( \frac{k \pi }{5n} \right) \\= & {} 2C(m,n/2) - C(m,n), \end{aligned}$$

where n can only be even. By following (9.2) we can express the above result as

$$\begin{aligned}&\frac{1}{5} \sum _{k=0}^{5n-1} \cos \left( \frac{\pi k}{5}\right) \left( 2 \cos \left( \frac{2 \pi k}{5} \right) + 2 \cos \left( \frac{4 \pi k}{5} \right) +1 \right) \cos ^{2m} \left( \frac{k \pi }{5n} \right) \\&\quad =2C(m,n/2) - C(m,n). \end{aligned}$$

After a little algebra we arrive at

$$\begin{aligned}&\sum _{k=0}^{5n-1} \left( 2 \cos \left( \frac{3 \pi k}{5} \right) + 2 \cos \left( \frac{ \pi k}{5} \right) \right) \cos ^{2m} \left( \frac{k \pi }{5n} \right) \\&\quad = \, 10C(m,n/2) - 2 C(m,5n/2) + C(m,5n)- 5 C(m,n). \end{aligned}$$

Once again, we are unable to decouple the component sums. In fact, as one goes to higher primes, there will be more component sums appearing in the final result, which makes the task of isolating them on their own even more difficult to accomplish. Nevertheless, we can combine the cosines on the lhs, thereby obtaining

$$\begin{aligned}&\sum _{k=0}^{5n-1} \cos \left( \frac{\pi k}{5} \right) \cos \left( \frac{ 2\pi k}{5} \right) \cos ^{2m} \left( \frac{k \pi }{5n} \right) \nonumber \\&= \frac{1}{4} \left( 10 C(m,n/2) - 2 C(m,5n/2) + C(m,5n)- 5 C(m,n) \right) . \end{aligned}$$

Comparing the above result with (9.6), we see that they are the alternating versions of one another.

We can, however, derive a result for the first sum on the rhs of (9.4), although it may not be regarded as very elegant. First, we express the sum as

$$\begin{aligned} \sum _{k=0}^{5n-1} \cos \left( \frac{2 k \pi }{5} \right) \cos ^{2m} \left( \frac{k \pi }{5n} \right) = \sum _{k=0}^{5n-1} \cos \left( \frac{2kn \pi }{5n} \right) \cos ^{2m} \left( \frac{k \pi }{5n} \right) . \end{aligned}$$

(9.7)

From [27, No. I.1.10], we have

$$\begin{aligned} \cos \left( \frac{2n k\pi }{5n} \right)= & {} 2^{2n-1} \cos ^{2n} \left( \frac{k \pi }{5n} \right) +\, n \sum _{j=0}^{n-1} \frac{(-1)^{j+1}}{j+1}\, \left( {\begin{array}{c}2n-j-2\\ j\end{array}}\right) 2^{2n-2j-2}\nonumber \\&\times \,\cos ^{2n-2j-2} \left( \frac{k \pi }{5n} \right) . \end{aligned}$$

(9.8)

Introducing (9.8) into (9.7) yields

$$\begin{aligned}&\sum _{k=0}^{5n-1} \cos \left( \frac{2 k \pi }{5} \right) \cos ^{2m} \left( \frac{k \pi }{5n} \right) \\&\quad = \sum _{k=0}^{5n-1} \left( 2^{2n-1} \cos ^{2m+2n} \left( \frac{k \pi }{5n} \right) + n \sum _{j=0}^{n-1} \frac{(-1)^{j+1}}{j+1}\, 2^{2n-2j-2} \right. \nonumber \\&\qquad \left. \times \left( {\begin{array}{c}2n-j-2\\ j\end{array}}\right) \cos ^{2m+2n-2j-2} \left( \frac{k \pi }{5n} \right) \right) . \end{aligned}$$

Recognizing that the sum over k is the basic cosine power sum defined in Sect. 2, we finally arrive at

$$\begin{aligned} \sum _{k=0}^{5n-1} \cos \left( \frac{2 k \pi }{5} \right) \cos ^{2m} \left( \frac{k \pi }{5n} \right) =&\; 2^{2n-1}\, C(m+n,5n) +n \sum _{j=0}^{n-1} \frac{(-1)^{j+1}}{j+1}\; 2^{2n-2j-2} \nonumber \\&\times \, \left( {\begin{array}{c}2n-j-2\\ j\end{array}}\right) \, C(m+n-j-1,5n). \end{aligned}$$

(9.9)

The other sum involving $\cos (4 k \pi /5)$ instead of $\cos (2 k \pi /5)$ can be directly obtained from (9.4). Although (9.9) is cumbersome, it does nevertheless demonstrate that the basic cosine power sum given above is combinatorial in nature or rational as a consequence of Theorem 2.1.