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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.

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    $\begingroup$ Wow, this is really horrible. $\endgroup$ Dec 15, 2018 at 21:25
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    $\begingroup$ Have you tried replacing the cosine by an complex exponential, using the Newton formula and studying the different factors? $\endgroup$
    – Aphelli
    Dec 15, 2018 at 21:36
  • $\begingroup$ @Mindlack Which formula by Newton? $\endgroup$
    – Gabriel
    Dec 15, 2018 at 22:12
  • $\begingroup$ I meant $(1+z)^n=\ldots$. $\endgroup$
    – Aphelli
    Dec 15, 2018 at 22:22
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    $\begingroup$ Factor out everything that does not depend on $k$. $\endgroup$
    – Aphelli
    Dec 15, 2018 at 22:29

1 Answer 1

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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 $$\lim_{L\to\infty} \cos\left(\frac{2\pi\ell\sqrt{\tau}}{L}\right)^{L^2}\cos(2\pi\ell x)=\exp(-2\pi^2\ell^2\tau)\cos(2\pi\ell x).$$

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