0
$\begingroup$

Is it possible to study the behavior of the floor function at infinity by estimating its growth?

The floor function has countably many discontinuities at integers, so I'm afraid that these discontinuities will somehow make it hard to study its behavior.

We all have seen in a first calculus course that $$\lim_{n \to \infty} \lfloor n \rfloor = n + \text{ other negligible terms }$$

But is it possible to make this estimation more accurate as $n \to \infty$?

The reason that I'm asking this is because of this post of mine in here.

I'm trying to show that the following limit exists and is equal to $r$:

$$\lim_{n \to \infty} \sqrt{\lfloor (n+r)^2 \rfloor+1}-\sqrt{n^2}$$

I've already shown that:

$$\lim_{n \to \infty} \sqrt{\lfloor (n+r)^2 \rfloor+1}-\sqrt{n^2} = \lim_{n \to \infty}\frac{\lfloor (n+r)^2 \rfloor+1-n^2}{\sqrt{\lfloor (n+r)^2 \rfloor+1}+\sqrt{n^2}} = \lim_{n \to \infty} \frac{\lfloor (n+r)^2 \rfloor - n^2}{2n}$$

But now I'm stuck because now I need an asymptotic expansion of $\lfloor x \rfloor$ at infinity that gives better estimates.

Any ideas are appreciated.

$\endgroup$
  • $\begingroup$ Try using the fact that $$\forall x \, \big(x-1 \leq \lfloor x \rfloor \leq x\big).$$ $\endgroup$ – Unwisdom Mar 13 '14 at 19:48
  • 2
    $\begingroup$ Your first equality is wrong and the limit is $+\infty$. $\endgroup$ – user63181 Mar 13 '14 at 19:51
  • $\begingroup$ @SamiBenRomdhane: The limit is not $+\infty$ unless I have made some typos. Use wolfram alpha and you'll see that it DOES converge. $\endgroup$ – math.n00b Mar 13 '14 at 19:54
  • 1
    $\begingroup$ @SamiBenRomdhane: Ah, now I see what you mean. You meant $\lim_{n \to \infty} \lfloor n \rfloor = n$. I thought you were referring to the limit in problem. Yes, I meant that as a limit equality, not an actual equality. I'll edit my post. $\endgroup$ – math.n00b Mar 13 '14 at 20:05
  • 1
    $\begingroup$ @dadexix86: Yeah, just now I understood what he meant. Pardon me. $\endgroup$ – math.n00b Mar 13 '14 at 20:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.