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This is along the lines of Problem 9.8. in 'Concrete Mathematics' by Graham, Knuth and Patashnik.

Does any of the relation $\prec$, $\succ$ or $\sim$ exist between functions $f(n) =\displaystyle \sum_{k=0}^{n}k^{\lfloor \cos (k) \rfloor}$ and $g(n) =n^{\frac{3}{2}}$?

Both definitely diverge monotone to infinity, but I can't get my head around the rest.

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    @sigma: the summation be over $n$?2011-05-31
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    @sigma: Also, do you mean to write $f(n) = \displaystyle\sum_{k=0}^n k^{\lfloor \cos k \rfloor}$?2011-05-31
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    You may not be imitating the problem closely enough. I am not able to guess what is intended, perhaps $\sum_{k=0}^n k^{\lfloor\cos k\rfloor}$.2011-05-31
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    sorry about this, corrected to sum over $n$. the function in the power of $n$ is floor(cos(n))2011-05-31
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    $\llcorner \cos (n) \lrcorner$ is either 0 or $-1$, and if I am not mistaken among any 6 consecutive $n$ at least t2o and at most four have $\llcorner \cos (n) \lrcorner=-1$. This should give good lower/upper bounds for your sum....2011-05-31
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    @user6312: it's not quite an imitation of the problem...2011-05-31
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    @sigma.z.1980: $f$ as you have written it does not depend on $n$, it is "constant" sort of, actually infinite, or more properly does not exist.2011-05-31
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    @user6312: it's just a shorthand notation. Of course it is a function of $n$2011-05-31
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    @user9176: why can't it be equal to 1?2011-05-31
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    Because the notation you use is standard for the floor function (integer part), and $\cos(n)\neq 1$ for any nonzero integer. If you mean the closest integer function, then you should specify it in the problem2011-05-31
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    no, I mean for all integer $n$. Is there some proof that $\cos(n) \neq 1$?2011-05-31
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    It is a standard fact that $\pi$ is irrational. (I am assuming that as usual by $\cos(x)$ you mean the cosine function, where $x$ is measured in **radians**.)2011-05-31

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Since $\cos k \in (-1,1)$ for any positive integer $k$, $\left\lfloor {\cos k} \right\rfloor \in \{ 0, - 1\}$. Hence, $$ f(n) = \sum\limits_{k = 0}^n {k^{\left\lfloor {\cos k} \right\rfloor } } \le \sum\limits_{k = 1}^n {k^0 } = n. $$

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    @Shai: What if $\lfloor \cos k \rfloor$ is replaced by $\lceil \cos k \rceil$?2011-05-31
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    @DJC: Good point, that would make the problem interesting (but this is not what OP asked).2011-05-31
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    It is not hard to see that $\cos k$ is often positive, since we are advancing by about $60$ degrees each time. That forces the altered version of $f$ to lie between $an^2$ and $bn^2$ for positive constants $a$ and $b$.2011-05-31
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    @Shai: Thank you. @user6312: $1$ radian is about $57^{\circ}$, but I would think that $\{\cos k\}_{k \in \Bbb{N}}$ would be pretty equidistributed. I may think about this more, and possibly start a new thread for this new question.2011-05-31
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    @DJC: Perhaps, ping only works for the first @: http://meta.math.stackexchange.com/questions/2063/ping-only-works-for-the-first2011-05-31
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    @user6312: 1 radian is about 57∘, but I would think that {cosk}k∈N would be pretty equidistributed. I may think about this more, and possibly start a new thread for this new question.2011-05-31
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    @Shai. Thank you again. I guess I should have known that by now.2011-05-31
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    I was mentioning the $an^2$ and $bn^2$ in the context of comparison with $n^{3/2}$, since these kinds of bounds are easy to establish, and enough. But a stronger reasonable conjecture is that if $f$ is the altered function, then $\lim_{n\to\infty} \frac{f(n)}{n^2}$ exists.2011-05-31
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    @DJC: user6312 added a comment (the one above).2011-05-31
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    @Shai: OK, but if $n \to \infty$, then $f(n)$ is undefined sine $\cos(\infty)$ is undefined and $g(n) \to \infty$ and so these relations cannot be established.2011-05-31
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    @sigma.z.1980: $f(n)$ is defined for any positive integer $n$, and is bounded from above by $n$. This is the only thing that matters here, and you certainly shouldn't set $n=\infty$. That's the idea with limits, in general.2011-05-31
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    OK, then in such case it tends to infinity at the order of $n$, which is of course slower than $n^{\frac{3}{2}}$2011-05-31
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    @sigma.z.1980: Yes (more precisely, $f(n)=O(n)$ as $n \to \infty$).2011-05-31