I would like to know if there a closed form solution for the sum:

$$ S_n(t) = \sum_{k=0}^{n} \cos( t \sqrt{k} ) $$

There is obviously an easy answer when the sum is replaced by an integral so this question is really asking for the exact form that $S_n(t)$ takes.

Any hints or references on how to approach this problem would be welcome.

  • $\begingroup$ Have you tried a Taylor expansion? The coefficients should be readily found. $\endgroup$ – Raskolnikov Jan 26 '11 at 22:31
  • $\begingroup$ @Raskolnikov: I get $ d^{2v} S_n(0) / dt^{2v} = (-1)^v \sum_{k=0}^n k^v $ (odd values of $v$ produce $\sin$ terms which zero out when evaluated at 0), which means $S_n(x) = \sum_{m=0}^{\infty} (-1)^m H_{n,m} x^m / m! $, where $H_{n,m}$ is the harmonic number in $n$ and $m$. WolframAlpha doesn't know how to evaluate this (nor do I). Any suggestions? $\endgroup$ – user4143 Jan 26 '11 at 23:07
  • 1
    $\begingroup$ What would suggest to you that a closed form exists? $\endgroup$ – Qiaochu Yuan Jan 26 '11 at 23:08
  • $\begingroup$ If the Taylor series expansion is not known by WolframAlpha, chances are that there is no closed form expression. There are formulas for the harmonic numbers, maybe they can help, but I guess they will at best allow you to reexpress $S_n(t)$ in terms of other sums of functions. $\endgroup$ – Raskolnikov Jan 26 '11 at 23:11
  • 1
    $\begingroup$ @user4143: there is an easy way to tell if a closed form exists: find one. However, there is no easy way to tell if a closed form does not exist. There are ways, I think, but they are hard, and I am not familiar with them, and they probably don't always work. The best you can do is probably to show that if a closed form exists then it has bad properties. $\endgroup$ – Qiaochu Yuan Jan 27 '11 at 9:24

$S_{n}=\sum^{n}_{k=0}\cos(t\sqrt{k})=\sum^{n}_{k=0}\frac{e^{it\sqrt{k}}+e^{-it\sqrt{k}}}{2}$. Any closed form of this must somehow evaluate $\sum^{n}_{k=0}e^{it\sqrt{k}}$. Consider $e^{x}=\sum^{\infty}_{i=0}x^{i}/i!$, $e^{it\sqrt{k}}$ would be $\sum^{\infty}_{j=0}(it\sqrt{k})^{j}/j!$. Hence the series would be:


This can be decomposed into even and odd $j$ terms. Since $j$ evaluate from $0$ we name the even one to be $S_{0}$ and the odd one $S_{1}$. The even terms $j=2s$ sums up to:


And the odd one $j=2s+1$ sums up to:


Now, what is the closed form of the original series? It is the same as $+_{0}(t)$. Switch $k$ and $s$ we should have $$\sum^{\infty}_{s=0}\sum^{\infty}_{k=0}-[(t^{2s})/(2s)!]k^{s}$$

I think there is NO such closed form to evaluate this double series.

  • 1
    $\begingroup$ Quite a long-winded answer, but finally you come to the correct answer to the question: "No". $\endgroup$ – GEdgar May 7 '11 at 13:36

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.