Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
@sigma: the summation be over $n$? –  user17762 May 31 '11 at 4:33
3  
@sigma: Also, do you mean to write $f(n) = \displaystyle\sum_{k=0}^n k^{\lfloor \cos k \rfloor}$? –  JavaMan May 31 '11 at 4:38
3  
You may not be imitating the problem closely enough. I am not able to guess what is intended, perhaps $\sum_{k=0}^n k^{\lfloor\cos k\rfloor}$. –  André Nicolas May 31 '11 at 4:38
1  
@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. –  André Nicolas May 31 '11 at 4:49
1  
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.) –  André Nicolas May 31 '11 at 5:10

1 Answer 1

up vote 2 down vote accepted

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

share|improve this answer
    
@Shai: What if $\lfloor \cos k \rfloor$ is replaced by $\lceil \cos k \rceil$? –  JavaMan May 31 '11 at 5:15
    
@DJC: Good point, that would make the problem interesting (but this is not what OP asked). –  Shai Covo May 31 '11 at 5:39
    
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 $b$. –  André Nicolas May 31 '11 at 5:47
    
@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. –  JavaMan May 31 '11 at 5:56
    
@DJC: Perhaps, ping only works for the first @: meta.math.stackexchange.com/questions/2063/… –  Shai Covo May 31 '11 at 6:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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