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My brain had twisted because of this nasty problem.

let

$$r_{n}=\sqrt{n^2+n+\frac{3}{4}}$$

$$x_{n}=\left \lfloor \frac{r_{n}}{\sqrt{2}}-\frac{1}{2} \right \rfloor$$

$$a_{n}=\sum_{k=1}^{\left \lfloor r_{n}-x_{n} \right \rfloor} \left \lfloor \sqrt{n^2+n-k^2-k+\frac{1}{2}-(x_{n})^2-(2k+1)x_{n}}-\frac{1}{2} \right \rfloor $$

$$A_{n}=4 ( (x_{n})^2+2a_n+n )+1$$

Question. How can I find the limit of below?

$$\lim_{n\to\infty}\frac{A_{n}}{n^2}$$

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    $\begingroup$ It looks like $A_n$ counts the lattice points in an ellipse or something. Try to draw a figure and interprete the occurring quantities therein. $\endgroup$ Jul 23, 2011 at 9:02

2 Answers 2

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As Artem has noted in the previous answer, $r_n\sim n$, $x_n\sim n/\sqrt2$ and $\lfloor r_n-x_n\rfloor\sim(1-1/\sqrt2)n$. Then $$ \frac{a_n}{n^2}\sim\frac{1}{n}\sum_{k=1}^{\lfloor(1-1/\sqrt2)n\rfloor}\sqrt{\frac{1}{2}-\sqrt2\,\frac{k}{n}-\Bigl(\frac{k}{n}\Bigr)^2} $$ which is a Riemann sum for the integral $$ \int_0^{1-1/\sqrt2}\sqrt{\frac{1}{2}-\sqrt2\,x-x^2}\,dx=\frac{\pi-2}{8}=0.142669\dots $$ This agrees very well with numerical computations; the relative error for $n=2^{20}$ is $3.1\times10^{-6}$. Finally $$ \lim_{n\to\infty}\frac{A_n}{n^2}=4\Bigl(\frac12+\frac{\pi-2}{4}\Bigr)=\pi. $$

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  • $\begingroup$ @kso830 Sorry, I misread the question. I have edited my answer. $\endgroup$ Jul 24, 2011 at 14:13
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Let's use some little-O notation and hand-waving:

$$r_n = n + o(n)$$

$$x_n = \frac{n}{\sqrt{2}} + o(n)$$

$$\lfloor r_n - x_n\rfloor = \frac{2 - \sqrt{2}}{2}n + o(n)$$

$$a_n = \sum_{k = 1}^{\frac{2 - \sqrt{2}}{2}n + o(n)} \Big(\sqrt{1 - \frac{1}{\sqrt{2}}}\Big) n + o(n) - k + o(k) - \sqrt{2kn} $$

Thus,

$$ a_n = \Big(\frac{2 - \sqrt{2}}{2}\Big)^{3/2} n^2 + o(n^2) - \Big(\frac{2 - \sqrt{2}}{4} \Big) n^2 + o(n^2) - \sqrt{2n} H_{\frac{2 - \sqrt{2}}{2} n + o(n),-1/2} $$

Unfortunately I can only bound the last term:

$$ -\Big(\frac{2 - \sqrt{2}}{4}\Big) n^2 \leq a_n - o(n^2) \leq \Bigg(\Big(\frac{2 - \sqrt{2}}{2}\Big)^{5/2} - \frac{2 - \sqrt{2}}{4}\Bigg) n^2 $$

$$ - 0.14645 n^2 \leq a_n - o(n^2) \leq -0.1 n^2$$

Thus, we get:

$$ 0.8284 \leq \lim_{n \rightarrow \infty} \frac{A_n}{n^2} \leq 1.2$$

Hope that helps if Christian Blatter's comment doesn't get you the answer already.

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  • $\begingroup$ There is probably a mistake when I count the summation from the first form of $a_n$ to the second. I will double check it again later... I am definitely not an expert on thus, it was just the first thing that popped into my head. $\endgroup$ Jul 24, 2011 at 6:22

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