TITLE: Observation: $\big(\frac{\sin x }{x}\big) ^n \sim e^{-a_n x^2}$
QUESTION [5 upvotes]: I have observed that this 
$$\big(\frac{\sin x }{x}\big) ^n \sim e^{-a_n x^2}$$

(It can be fitted better than this, but I did not want them to completely overlap)
I want to learn how I would find $a_n$, or if this relationship is technically asymptotic to begin with.

REPLY [4 votes]: You can understand this using probability theory.
The function $x\mapsto{\sin(x)\over x}$ is the characteristic function of a random variable $X$ that is uniformly distributed over $(-1,1)$. This random variable has mean $\mu=0$ and variance $\sigma^2=1/3$. By the central limit theorem, the normalized sum ${1\over \sqrt{n}}\sum_{i=1}^n X_i$ has a Gaussian limit, so 
$$\left[{\sin(x/\sqrt{n})\over (x/\sqrt{n})}\right]^n\to \exp(-x^2\sigma^2/2), $$
or
$$\left[{\sin(x)\over x}\right]^n\approx \exp(-n x^2/6)$$
for small $x$.