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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.

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Have you tried a Taylor expansion? The coefficients should be readily found. –  Raskolnikov Jan 26 '11 at 22:31
    
@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? –  user4143 Jan 26 '11 at 23:07
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What would suggest to you that a closed form exists? –  Qiaochu Yuan Jan 26 '11 at 23:08
    
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. –  Raskolnikov Jan 26 '11 at 23:11
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@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. –  Qiaochu Yuan Jan 27 '11 at 9:24
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1 Answer 1

$S_{n}=\sum^{n}_{k=0}\cos(t\sqrt{k})=\sum^{n}_{k=0}\frac{e^{it\sqrt{k}}+e^{-it\sqrt{k}}}{2}$. 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:

$+_{n}(t)=\sum^{\infty}_{k=0}\sum^{\infty}_{j=0}(it\sqrt{k})^{j}/j!$.

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:

$+_{0}(t)=\sum^{\infty}_{k=0}\sum^{\infty}_{s=0}-(t^{2s}k^{s})/(2s)!$.

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

$+_{1}(t)=-i\sum^{\infty}_{k=0}\sum^{\infty}_{s=0}(t\sqrt{k})^{2s+1}/(2s+1)!$

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.

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