# Horrible limit envolving floor function

Let $$x\in [0,1]$$, $$\ell\in\mathbb{Z}$$ and $$\tau>0$$. I want to calculate $$\lim_{L\to\infty}\sum_{k=0}^{\lfloor \tau L^2\rfloor}\frac{1}{2^{\lfloor \tau L^2\rfloor}}\binom{\lfloor \tau L^2\rfloor}{k}\cos\left(2\pi \ell\frac{\lfloor xL\rfloor-\lfloor \tau L^2\rfloor+2k}{L}\right).$$

I think the result is $$\exp(-2\pi^2\ell^2\tau)\cos(2\pi\ell x)$$ but I have no idea in how to prove it.

I tried estimating the binomial with Stirling's aproximation but without success.

• Wow, this is really horrible. – thesagniksaha Dec 15 '18 at 21:25
• Have you tried replacing the cosine by an complex exponential, using the Newton formula and studying the different factors? – Mindlack Dec 15 '18 at 21:36
• @Mindlack Which formula by Newton? – Gabriel Dec 15 '18 at 22:12
• I meant $(1+z)^n=\ldots$. – Mindlack Dec 15 '18 at 22:22
• Factor out everything that does not depend on $k$. – Mindlack Dec 15 '18 at 22:29

You are correct. We have that \begin{aligned} & \sum_{k=0}^{\lfloor \tau L^2\rfloor}\frac{1}{2^{\lfloor \tau L^2\rfloor}}\binom{\lfloor \tau L^2\rfloor}{k}\cos\left(2\pi \ell\frac{\lfloor xL\rfloor-\lfloor \tau L^2\rfloor+2k}{L}\right) \\ = &\sum_{k=0}^{\lfloor \tau L^2\rfloor}\frac{1}{2^{\lfloor \tau L^2\rfloor}}\binom{\lfloor \tau L^2\rfloor}{k}\Re\left(\exp(i2\pi\ell \frac{\lfloor xL\rfloor-\lfloor \tau L^2\rfloor+2k}{L})\right) \\ =& \frac{1}{2^{\lfloor \tau L^2\rfloor}}\Re\left(\sum_{k=0}^{\lfloor \tau L^2\rfloor}\binom{\lfloor \tau L^2\rfloor}{k}\exp(i2\pi\ell \frac{\lfloor xL\rfloor-\lfloor \tau L^2\rfloor+2k}{L})\right) \\ =& \frac{1}{2^{\lfloor \tau L^2\rfloor}}\Re\left(\exp\left(i2\pi\ell\frac{\lfloor xL\rfloor -\lfloor\tau L^2\rfloor}{L}\right)\left(1+\exp\left(i\frac{4\pi\ell}{L}\right)\right)^{\lfloor \tau L^2\rfloor}\right). \end{aligned} Since we are taking the limit as $$L\to\infty$$ we have that $$\frac{\lfloor xL\rfloor -\lfloor\tau L^2\rfloor}{L}\to x-\tau L$$ so we can consider the expression $$\frac{1}{2^{\lfloor \tau L^2\rfloor}}\Re\left(\exp\left(i2\pi\ell(x-\tau L)\right)\left(1+\exp\left(i\frac{4\pi\ell}{L}\right)\right)^{\lfloor \tau L^2\rfloor}\right).$$ Again since we only care about $$L\to\infty$$ we can let $$L\mapsto L/\sqrt{\tau}$$ so that we are considering the expression \begin{aligned} &\frac{1}{2^{L^2}}\Re\left(\exp\left(i2\pi\ell(x-\sqrt{\tau} L)\right)\left(1+\exp\left(i\frac{4\pi\ell\sqrt{\tau}}{L}\right)\right)^{L^2}\right) \\ =&\Re\left(\exp(i2\pi\ell x)\left(\frac{\exp\left(-i\frac{2\pi\ell\sqrt{\tau}}{L}\right)+\exp\left(i\frac{2\pi\ell\sqrt{\tau}}{L}\right)}{2}\right)^{L^2}\right) \\ =&\cos\left(\frac{2\pi\ell\sqrt{\tau}}{L}\right)^{L^2}\Re\left(\exp(i2\pi\ell x)\right) \\ =& \cos\left(\frac{2\pi\ell\sqrt{\tau}}{L}\right)^{L^2}\cos(2\pi\ell x). \end{aligned} Taking the limit we indeed get $$\lim_{L\to\infty} \cos\left(\frac{2\pi\ell\sqrt{\tau}}{L}\right)^{L^2}\cos(2\pi\ell x)=\exp(-2\pi^2\ell^2\tau)\cos(2\pi\ell x).$$