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I want compute, in a closed form or an asymptotic (with a, big oh as, error term) this mean

$$\delta_k(n):=\sum_{m=0}^{k-1}\cos(\frac{2\pi m n}{k \log 2})$$

defined for each integer $k\geq 1$. Truly I am interesting in $\sum_{1\leq k\leq n}\delta_k(n)$ in this exercise, but if your computations are good, it should be satisfactory.

I've made some humble computations (the good analysis that solve this exercise is required for you) using Euler-MacLaurin summation with $P_1(x)=x-\lfloor x\rfloor-\frac{1}{2}$, looking to compute $$\delta_{k}(n)=1+\sum_{m=1}^{k-1}cos(\frac{2\pi m n}{k \log 2}).$$ Now,

$$\sum_{m=1}^{k-1}\cos(\frac{2\pi m n}{k \log 2})=\int_{1}^{k-1}\cos(\frac{2\pi x n}{k \log 2})dx-\frac{2\pi n}{k\log 2}\int_1^{k-1}P_1(x)\sin(\frac{2\pi x n}{k \log 2})dx$$ $$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+\frac{1}{2}(\cos(\frac{2\pi n}{k \log 2})+\cos(\frac{2\pi (k-1) n}{k \log 2})).$$

And $$\int_{1}^{k-1}\cos(\frac{2\pi x n}{k \log 2})dx=\frac{k\log 2}{2\pi n}\left(\sin(\frac{2\pi (k-1) n}{k \log 2})-\sin(\frac{2\pi n}{k \log 2})\right).$$

Too, I know that $x-\lfloor x\rfloor=O(1)$, and by MacLaurin series expansion for the sine function that $\sin (\frac{2\pi x n}{k \log 2})=\frac{2\pi x n}{k \log 2}+O((\frac{2\pi x n}{k \log 2})^3)$.

Then, (if my computations are rigth)

Question. Can you give a closed form or an asymptotic for $\delta_k(n)$, as $k\to\infty$? Thanks in advance.

You can ask to me about more context, but is only a curiosity build as my previous post.

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    $\begingroup$ There are closed forms for sums of cosine over any sequence $x=a+kb$ (an "arithmetic sequence" of inputs to the cosine). That may simplify your initial $\delta_k(n)$ so it no longer contains $k$ but it will have $n$ still in it. $\endgroup$ – coffeemath Dec 12 '15 at 19:21
  • $\begingroup$ Very thanks much @coffemath, I hope that the solver of my question uses your good hint. I take notes and try, but I will wait a full answer. Very thanks much, I vote up your comment , as useful. I love coffe too, more much than you loves coffe! $\endgroup$ – user243301 Dec 12 '15 at 19:26
  • $\begingroup$ I should have said "simplify so that it no longer has $m$ in it" for your case, since $m$ is what is summed over. It will still have $k$ in it since the upper limit on the sum is $k-1.$ $\endgroup$ – coffeemath Dec 12 '15 at 20:17
  • $\begingroup$ Thanks, my english and abilities aren't the best. Very thanks much @coffeemath $\endgroup$ – user243301 Dec 12 '15 at 22:29
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Let $x=2\pi n/\log(2)$ and let $f_k(x)$ be the sequence given by

$$f_k(x)=\sum_{m=0}^{k-1}\cos (m\,x/k) \tag 1$$

Then, we can write the sum in $(1)$ as

$$\begin{align} \sum_{m=0}^{k-1}\cos (m\,x/k) &=\text{Re}\left(\sum_{m=0}^{k-1}e^{i m\,x/k} \right)\\\\ &=\text{Re}\left(\frac{1-e^{ix}}{1-e^{ix/k}}\right)\\\\ &=\sin^2(x/2)+\frac12 \sin(x)\,\cot(x/2k) \end{align}$$

Next, we expand the cotangent function and write

$$f_k(x)=\frac{2\sin(x)\,k}{x}+\sin^2(x/2)-\frac{x\sin(x)}{6k}-\frac{x^3\sin(x)}{360k^3}+O\left(\frac{1}{k^5}\right)$$

Therefore, for $\delta_k(n)$ we have

$$\begin{align} \delta_k(n)&=\left(\frac{\log (2)\sin\left(\frac{2\pi n}{\log(2)}\right)}{n\pi}\right)k\\\\ &+(1+\sin^2\left(n\pi/\log(2)\right))\\\\ &-\left(\frac{n\pi\,\sin\left(\frac{2\pi n}{\log(2)}\right)}{3\log(2)}\right)\frac1k\\\\ &+\left(\frac{(n\pi)^3\sin\left(\frac{2\pi n}{\log(2)}\right)}{45\log^3(2)}\right)\frac{1}{k^3}\\\\ &+O\left(\frac{1}{k^5}\right) \end{align}$$

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  • $\begingroup$ Uff, incredible, I vote up your answer, tomorrow I see the computations and choose your answer, now is very later here to think. $\endgroup$ – user243301 Dec 12 '15 at 22:30
  • $\begingroup$ Pleased to hear this was useful! -Mark $\endgroup$ – Mark Viola Dec 13 '15 at 3:40
  • $\begingroup$ Very thanks much to you @Dr.MV, this was only a puzzle but I write you that this sum, $\delta_k(n)$ if there are no mistakes is $$\sum_{m=0}^{k-1}\frac{\Re (e^{\frac{ms_n}{k}})}{e^{\frac{m}{k}}},$$ where $s_n$, for $n\geq 1$ are the zeros of Dirichlet eta function of the form $s_n=1+\frac{2\pi n}{\log 2}i$. As I've said is only a curiosity, has no sense mathematical. $\endgroup$ – user243301 Dec 13 '15 at 8:05
  • $\begingroup$ You're welcome. My pleasure $\endgroup$ – Mark Viola Dec 13 '15 at 17:40
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There is a known formula $$\sum_{m=0}^{k-1} \cos (m x)=\cos \frac{(k-1)x}{2} \sin \frac{k x}{2} \csc \frac{x}{2}.$$ In this put $$x=\frac{2\pi n}{k \log 2}$$ and it will match your $\delta_k(n).$

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  • $\begingroup$ Very thanks much @coffeemath, this was only a puzzle but I write you that this sum, $\delta_k(n)$ is $$\sum_{m=0}^{k-1}\frac{\Re (e^{\frac{ms_n}{k}})}{e^{\frac{m}{k}}},$$ where $s_n$, for $n\geq 1$ are the zeros of Dirichlet eta function of the form $s_n=1+\frac{2\pi n}{\log 2}i$. As I've said is only a curiosity, has no mathematical sense. $\endgroup$ – user243301 Dec 13 '15 at 7:58
  • $\begingroup$ I copy your maths and try think with your formula too, coffeemath. $\endgroup$ – user243301 Dec 13 '15 at 8:00

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