Función de límite horrible envolving piso

Question

Que <span class="math-container">$x\in [0,1]$</span>y <span class="math-container">$\ell\in\mathbb{Z}$</span> <span class="math-container">$\tau>0$</span>. Quiero calcular <span class="math-container">$$\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).$ $</span>

Creo que el resultado es <span class="math-container">$\exp(-2\pi^2\ell^2\tau)\cos(2\pi\ell x)$</span> pero no tengo ni idea de cómo probarlo.

He intentado calcular el binomio con aproximación de Stirling, pero sin éxito.

Answer 1

Estás en lo correcto. Tenemos que $$\begin{aligned} & \sum_{k=0}^{\lfloor \tau L^2\rfloor}\frac{1}{2^{\lfloor \tau L^2\rfloor}}\binom{\lfloor \tau L^2\rfloor}{k}\cos\left(2\pi \ell\frac{\lfloor xL\rfloor-\lfloor \tau L^2\rfloor+2k}{L}\right) \\ = &\sum_{k=0}^{\lfloor \tau L^2\rfloor}\frac{1}{2^{\lfloor \tau L^2\rfloor}}\binom{\lfloor \tau L^2\rfloor}{k}\Re\left(\exp(i2\pi\ell \frac{\lfloor xL\rfloor-\lfloor \tau L^2\rfloor+2k}{L})\right) \\ =& \frac{1}{2^{\lfloor \tau L^2\rfloor}}\Re\left(\sum_{k=0}^{\lfloor \tau L^2\rfloor}\binom{\lfloor \tau L^2\rfloor}{k}\exp(i2\pi\ell \frac{\lfloor xL\rfloor-\lfloor \tau L^2\rfloor+2k}{L})\right) \\ =& \frac{1}{2^{\lfloor \tau L^2\rfloor}}\Re\left(\exp\left(i2\pi\ell\frac{\lfloor xL\rfloor -\lfloor\tau L^2\rfloor}{L}\right)\left(1+\exp\left(i\frac{4\pi\ell}{L}\right)\right)^{\lfloor \tau L^2\rfloor}\right). \end{aligned}$$ Ya estamos tomando el límite cuando $L\to\infty$ tenemos que $$\frac{\lfloor xL\rfloor -\lfloor\tau L^2\rfloor}{L}\to x-\tau L$$ por lo que podemos considerar la expresión $$\frac{1}{2^{\lfloor \tau L^2\rfloor}}\Re\left(\exp\left(i2\pi\ell(x-\tau L)\right)\left(1+\exp\left(i\frac{4\pi\ell}{L}\right)\right)^{\lfloor \tau L^2\rfloor}\right).$$ De nuevo ya que nosotros sólo nos preocupamos de $L\to\infty$ podemos dejar $L\mapsto L/\sqrt{\tau}$ , de modo que estamos considerando la expresión $$\begin{aligned} &\frac{1}{2^{L^2}}\Re\left(\exp\left(i2\pi\ell(x-\sqrt{\tau} L)\right)\left(1+\exp\left(i\frac{4\pi\ell\sqrt{\tau}}{L}\right)\right)^{L^2}\right) \\ =&\Re\left(\exp(i2\pi\ell x)\left(\frac{\exp\left(-i\frac{2\pi\ell\sqrt{\tau}}{L}\right)+\exp\left(i\frac{2\pi\ell\sqrt{\tau}}{L}\right)}{2}\right)^{L^2}\right) \\ =&\cos\left(\frac{2\pi\ell\sqrt{\tau}}{L}\right)^{L^2}\Re\left(\exp(i2\pi\ell x)\right) \\ =& \cos\left(\frac{2\pi\ell\sqrt{\tau}}{L}\right)^{L^2}\cos(2\pi\ell x). \end{aligned}$$ Tomando el límite realmente llegamos $$\lim_{L\to\infty} \cos\left(\frac{2\pi\ell\sqrt{\tau}}{L}\right)^{L^2}\cos(2\pi\ell x)=\exp(-2\pi^2\ell^2\tau)\cos(2\pi\ell x).$$