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.

Not sure how to figure out this...I tried just doing normal $\frac{(x+\frac{1}{2})} {x}$ but I know that's not solving for the floor and I know I need a squeeze theorem..but not sure how to solve this one..thanks.

share|improve this question
1  
@Alex: Do you see how you've changed the potential meaning of the question? Also jtm22, if you're interested in the asymptotic behavior of this as $x\to\infty$ (as suggested in the chatroom), please divulge this information instead of leaving your readers in the dark. At any rate, if you are interested, consider dividing $x-1\le \lfloor x+1/2\rfloor\le x+1$ by $x$ and using squeeze theorem. –  anon Feb 20 '13 at 20:13
    
@jtm22: Can you clarify whether, when you wrote $\operatorname{floor}(x+1/2)/x$, you meant $\operatorname{floor}\big((x+1/2)/x\big)$ or $\big(\operatorname{floor}(x+1/2)\big)/x$? –  Rahul Feb 20 '13 at 20:38

3 Answers 3

$$ \left \lfloor \frac {x+\frac 12}x \right \rfloor = \left \lfloor 1 + \frac 1{2x} \right \rfloor = 1 + \left \lfloor \frac 1{2x} \right \rfloor = \left \{ \begin{array}{lcl} 1 & \text{if} & x \in \left (+\frac 12, +\infty \right ) \\ 0 & \text{if} & x \in \left (-\infty, -\frac 12 \right ] \\ m & \text{if} & m-1 \le \frac 1{2x} < m \\ -m & \text{if} & -m-1 \le \frac 1{2x} <-m \end{array}\right . $$ where $m \ge 1, m \in \mathbb N$

share|improve this answer

As $x-1\leq \operatorname{floor}(x+\frac{1}{2})\leq x+1 $ So you have $$ \frac{x-1}{x}=1-\frac{1}{x}\leq \frac{\operatorname{floor}(x+\frac{1}{2})}{x} \leq \frac{x+1}{x}=1+\frac{1}{x}$$

share|improve this answer

$$\left\lfloor\cfrac{x+\frac12}x\right\rfloor=\left\lfloor1+\frac1{2x}\right\rfloor=1+\left\lfloor\frac1{2x}\right\rfloor.$$

Note that $n=\left\lfloor\frac1{2x}\right\rfloor$ if and only if $n\le\frac1{2x}<n+1.$ Hence, $0\leq \frac1{2x}<1$ if and only if $1<2x$ if and only if $x>\frac12$, in which case $$\left\lfloor\cfrac{x+\frac12}x\right\rfloor=1+0=1.$$ Also, $-1\le\frac1{2x}<0$ if and only if $2x\le -1$ if and only if $x\le-\frac12$, in which case $$\left\lfloor\cfrac{x+\frac12}x\right\rfloor=1+-1=0,$$ so the end behavior is easy to calculate. If you want more detail, then for integers $n\neq0,-1$, we have $$n\leq\frac1{2x}<n+1$$ if and only if $$\frac1{n+1}<2x\le\frac1n$$ if and only if $$x\in\left(\frac1{2(n+1)},\frac1{2n}\right].$$ Hence, $$\left\lfloor\cfrac{x+\frac12}x\right\rfloor=\begin{cases}1 & x\in\left(\frac12,\infty\right),\\0 & x\in\left(-\infty,-\frac12\right],\\1+n & n\in\Bbb Z\smallsetminus\{-1,0\},x\in\left(\frac1{2(n+1)},\frac1{2n}\right].\end{cases}$$

share|improve this answer
    
in the chat she asked for the asymptotic behaviour of floor (x+1/2) –  Dominic Michaelis Feb 20 '13 at 20:19

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.