16
$\begingroup$

How to evaluate $\displaystyle\sum_{k=0}^{n-1} \sin^t\left(\frac{\pi k}{2n}\right)$?

$t,n$ are constants $\in \Bbb{Z}$.

My try:

$$\begin{align} \zeta:=e^{i\pi/2n} \implies & \sum_{k=0}^{n-1}\sin^t\left(\frac{\pi k}{2n}\right) \\ = &\frac{1}{2^t}\sum_{k=0}^{n-1}\left(\zeta^k- \zeta^{-k}\right)^{t} \\ = &\frac{1}{2^t}\sum_{k=0}^{n-1}\left(\zeta^{-kt}\right)\left(\zeta^{2k}-1\right)^{t}\\ \end{align}$$

How to proceed from here?

If a closed-form solution is not possible, can we still work out cases such as $t = n, t = 2n, $ etc?

EDIT: Sangchul Lee provided an answer for even t. Still looking for a solution for odd t as well.

$\endgroup$
  • $\begingroup$ Wouldn't $\zeta^{-kt}=i^{t/n}$? It seems the $n$ just disappears in that line... $\endgroup$ – Simply Beautiful Art Sep 18 '16 at 22:52
  • $\begingroup$ @SimpleArt you are correct, fixed $\endgroup$ – Teoc Sep 18 '16 at 22:54
  • $\begingroup$ Just an idea, but you can do binomial expansion on move the summations around. It didn't lead anywhere as far as I could see though: $$= \frac1{2^t}\sum_{\nu=0}^\infty\frac{(-1)^\nu t!}{\nu!(t-\nu)!}\frac{1-\left(\zeta^{t-2\nu}\right)^n}{1-\zeta^{t-2\nu}}$$ $\endgroup$ – Simply Beautiful Art Sep 18 '16 at 22:59
  • $\begingroup$ The title has $\sin ^t (\pi k / n )$ while in the text it is $\sin^t ( \pi k / 2n ) $, which one are you after? $\endgroup$ – anecdote Sep 25 '16 at 21:15
14
+25
$\begingroup$

Here is an answer when $t = 2s$ is even.


Let $n, s$ be non-negative integers. Then with $\zeta = \mathrm{e}^{i\pi/n}$ and $\omega = \zeta^2 = \mathrm{e}^{2i\pi/n}$, we have

\begin{align*} 2^{2s} \sum_{k=0}^{n-1} \sin^{2s} \left(\frac{\pi k}{n} \right) &= \sum_{k=0}^{n-1} \left( \frac{\zeta^k + \zeta^{n-k}}{i} \right)^{2s} \\ &= (-1)^{-s} \sum_{k=0}^{n-1}\sum_{l=0}^{2s} \binom{2s}{l} \zeta^{(2s-l)k+l(n-k)} \\ &= (-1)^{-s} \sum_{k=0}^{n-1}\sum_{l=0}^{2s} \binom{2s}{l} (-1)^l \omega^{(s-l)k}. \end{align*}

Interchanging the order of summation, we have

\begin{align*} 2^{2s} \sum_{k=0}^{n-1} \sin^{2s} \left(\frac{\pi k}{n} \right) &= \sum_{l=0}^{2s} (-1)^{l-s} \binom{2s}{l} \left( \sum_{k=0}^{n-1} (\omega^{s-l})^k \right) \\ &= \sum_{l=0}^{2s} (-1)^{l-s} \binom{2s}{l} \left( n \cdot \mathbf{1}_{\{ l \equiv s \ (\mathrm{mod} \ n)\}} \right) \\ &= n \sum_{j} (-1)^{nj} \binom{2s}{s+nj}, \end{align*}

where the last summation runs over all integers $j$ such that $-\frac{s}{n} \leq j \leq \frac{s}{n}$.

Examples. As special cases, plugging $s = n$ yields

$$ \sum_{k=0}^{n-1} \sin^{2n} \left(\frac{\pi k}{n} \right) = \frac{n}{2^{2n}} \left[ \binom{2n}{n} + (-1)^n 2 \right], $$

and similarly, replacing $n$ by $2n$ and plugging $s = n$ yields

$$ \sum_{k=0}^{n-1} \sin^{2n} \left(\frac{\pi k}{2n} \right) = \frac{n}{2^{2n}} \binom{2n}{n} - \frac{1}{2}. $$


Addendum. Here is an intuition on why we expect the sum to be simplified. If $t = 2s$ is even, then we can expand $\sin^{2s} x$ into a linear combination of $1, \cos 2x, \cos 4x, \cdots, \cos 2sx$. So in our case, we may write

$$ \sin^{2s} \left(\frac{\pi k}{n}\right) = \sum_{j=0}^{s} a_j \cos \left(\frac{2\pi j k}{n}\right). $$

Now, as you sum this over $k = 0, \cdots, n-1$, all the cosine terms will cancel out except when $j$ is a multiple of $n$. This means that

$$ \sum_{k=0}^{n-1} \sin^{2s} \left(\frac{\pi k}{n}\right) = n (a_0 + a_n + a_{2n} + \cdots). $$

So it is enough to identify $a_j$ for $j$'s multiple of $n$. This can be done by expanding $\sin^{2s} x$ using complex exponential and the binomial theorem. This is essentially what we did in the computation above.

On the other hand, this trick does not work for odd $t$. Indeed, when $t = 2s+1$ is odd, we can write

$$ \sin^{2s+1} \left(\frac{\pi k}{n}\right) = \sum_{j=0}^{s} a_j \sin \left(\frac{\pi (2j+1)k}{n}\right) $$

for some constants $a_0, \cdots, a_n$. (In fact, $a_j = (-1)^j 2^{-2s} \binom{2s+1}{s-j}$.) Now summing both sides for $k = 0, \cdots, n-1$,

\begin{align*} \sum_{k=0}^{n-1}\sin^{2s+1} \left(\frac{\pi k}{n}\right) &= \sum_{j=0}^{s} a_j \sum_{k=0}^{n-1} \sin \left(\frac{\pi (2j+1)k}{n}\right) \\ &= \frac{1}{2^{2s}} \sum_{j=0}^{s} (-1)^j \binom{2s+1}{s-j} \cot \left(\frac{\pi(2j+1)}{2n}\right). \end{align*}

It is hard for me to believe that this will ever simplify except for some nice $s$.


Addendum 2 (Just for fun). Although no longer simpler than the original sum, one can also prove that for complex $t$ with $\Re(t) > 0$ and for positive integer $n$ the following formula holds:

$$\sum_{k=0}^{n-1} \sin^{t} \left(\frac{\pi k}{n} \right) = \frac{n}{2^{t}} \sum_{j=-\infty}^{\infty} (-1)^{nj} \binom{t}{\frac{t}{2}+nj}. $$

Here, $\binom{n}{k} = \frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)}$ is the extended binomial coefficient.

$\endgroup$
8
$\begingroup$

This integral remind me of Matsubara sum. Let me use contour integral to provide another perspective of the case $t = 2s$.

Define $a_n = \sum_{k = 1}^{n-1} \sin ^{t} \left( \frac{\pi k }{n} \right) $. What OP is looking for is \begin{equation} \sum_{k=1}^{n-1} \sin^{t} \left( \frac{\pi k }{2n} \right) = \frac{1}{2} \left( \sum_{k=1}^{2n-1} \sin^{t} \left( \frac{\pi k }{2n} \right) -1 \right) = \frac{1}{2} ( a_{2n } - 1 ) \end{equation} One can extend the summation to all the roots of $\zeta^{2n} = 1$ if $t = 2s$ \begin{equation} a_n = \frac{1}{2} \sum_{k=1}^{2n} \sin^{2s} \left( \frac{\pi k }{n} \right ) = \frac{1}{2} \sum_{ k = 1}^{2n} f( \zeta_k ) \qquad \zeta_k = \exp( \frac{ik\pi }{n}) \end{equation} where \begin{equation} f(z ) = \left( \frac{z - \frac{1}{z}}{2i} \right)^{2s} \end{equation}

The summation can be regarded as the sum of the residue of the follow contour integral \begin{equation} \sum_{ k = 1}^{2n} f( \zeta_k ) = \frac{2n}{2\pi i }\oint_{C} \frac{z^{2n-1}}{ z^{2n} - 1} f( z) dz \end{equation} where the integrand is specifically designed such that \begin{equation} \text{Res} ( \frac{ 2n z^{2n-1}}{z^{2n} - 1} f(z) , \zeta_k ) = 2n \frac{z^{2n-1}(z - \zeta_k)}{z^{2n} - \zeta^{2n}_k} f(\zeta_k ) \Big|_{z = \zeta_k } = f(\zeta_k ) \end{equation} Here the closed contour should enclose all $\zeta_k$ but avoid the singularities of $f$, which can be taken as two concentric circles as shown in the following figure(red spots are those roots $\xi_k$).

enter image description here

The inner contour gives (minus) the residue at $0$, and the outer contour gives (minus) the residue at $\infty$, hence \begin{equation} \begin{aligned} a_n &= n \left[ -\text{Res}( \frac{z^{2n-1}}{ z^{2n} - 1} f( z) , 0 ) - \text{Res}( \frac{z^{2n-1}}{ z^{2n} - 1} f( z) , \infty ) \right] \\ \end{aligned} \end{equation} Convert the residue of $\infty$ to $0$ \begin{equation} \begin{aligned} a_n =& n \left[ \text{Res}( \frac{z^{2n-1}}{ 1 - z^{2n} } f( z) , 0 ) + \text{Res}( \frac{\frac{1}{z}}{1 - z^{2n} } f( \frac{1}{z}) , 0 ) \right] \\ =& n \left[ \text{Res}( \frac{1}{z}\frac{z^{2n} + 1 }{ 1 - z^{2n} } f(z) , 0 ) \right] \\ =& n \left[ \text{Res}( \frac{1}{z} ( 1 + 2\sum_{k=1}^{\infty} z^{2nj} ) f(z) , 0 ) \right] \\ \end{aligned} \end{equation} Finally, expand $f(z)$ and extract the terms with power $0, -2nj, -4nj$ etc, \begin{equation} \begin{aligned} a_n &= n \frac{(-1)^s}{2^{2s}} \left( (-1)^s {2s \choose s } + 2 \sum_{0 < nj < s } (-1)^{s+ nj} {2s \choose s+nj } \right)\\ &= \frac{n}{2^{2s}} \sum_{ -s < nj < s } (-1)^{nj} {2s \choose s + nj } \\ \end{aligned} \end{equation} This is the same as what @Sangchul Lee obtained.

$\endgroup$
0
$\begingroup$

I found this long ago:

$$\begin{aligned} {S_t}(n) &= \sum\limits_{k = 0}^n {{{\sin }^p}} (ak)\\ {\text{For odd p}}&:\\ {S_t}(n) &= \frac{1}{{{2^p}}}\sum\limits_{k = 0}^{\frac{{p - 1}}{2}} {{{( - 1)}^k}} C(p,\frac{{p - 1}}{2} - k){\text{Long}}\\ {\text{Long}} &= \left[ {\cot \left( {\frac{a}{2}(2k + 1)} \right) - \csc \left( {\frac{a}{2}(2k + 1)} \right)\cos \left( {\frac{a}{2}(4kn + 4k + 4n + 1)} \right)} \right]\\ {\text{For even p}}&:\\ {S_t}(n) &= \frac{1}{{{2^p}}}\left[ {\left( {\frac{1}{2} + n} \right)C(p,\frac{p}{2}) + \sum\limits_{k = 1}^{\frac{p}{2}} {{{( - 1)}^k}} C(p,\frac{p}{2} - k)\csc (ak)\sin (ak(2n + 1))} \right]\\ \end{aligned}$$

$C(n,k) = \left( \begin{gathered} n \\ k \\ \end{gathered} \right)$, just as a short to input $\LaTeX$.

I try to find my proof but in vain, and I am seeking for a new elegent proof.


So, for this question:

Let $p \to t,n \to n - 1,a \to \pi /2n$

It's:

$$\begin{aligned} {S_p}(n) &= \sum\limits_{k = 0}^{n - 1} {{{\sin }^t}} \left( {\frac{\pi }{{2n}}k} \right)\\ {\text{For odd n}}&:\\ {S_p}(n) &= {2^{ - t}}\sum\limits_{k = 0}^{\frac{{t - 1}}{2}} {{{( - 1)}^k}C(t,\frac{1}{2}( - 2k + t - 1)){\text{Long}}} \\ {\text{Long}} &= \csc \left( {\frac{{2\pi k + \pi }}{{4n}}} \right)\left( {\sin \left( {\frac{{\pi (k(4n - 2) - 1)}}{{4n}}} \right) + \cos \left( {\frac{{2\pi k + \pi }}{{4n}}} \right)} \right)\\ {\text{For even n}}&:\\ {S_p}(n) &= {2^{ - t}}\left( {\left( {n - \frac{1}{2}} \right)C(t,\frac{t}{2}) + \sum\limits_{k = 1}^{\frac{t}{2}} {{{( - 1)}^k}} C(t,\frac{t}{2} - k){\text{Long}}} \right)\\ {\text{Long}} &= \sin \left( {\frac{{\pi (k(2(n - 1) + 1))}}{{2n}}} \right)\csc \left( {\frac{{\pi k}}{{2n}}} \right)\\ \end{aligned}$$

It we want to find some nice closed-form , the $\text{Long}$ part should be simple.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.