on a quantisation of the bell curve

Question

The bell curve function: $e^{-x^2/2}$ is an eigenfunction of the Fourier transform (FT) on the real line. Is its quantisation/discretisation the binomial distribution (coefficients $n$ choose $k$) an eigenvector of the discrete FT? The question pertains to an EIT technique.

Answer 1

Let $F_{n+1}$ be the $(n+1)\times(n+1)$ Fourier matrix.

Let $C_n$ be the column vector whose coefficients are constituted by the $n$th line of Pascal's triangle.

The image of $C_n$ by $F_{n+1}$ is not proportional to $C_n$ but to a vector whose entries' moduli are proportional to the $n$th power of a cosine function $A cos^n(...)$.

Explicitly: the $L$th entry of vector $F_{n+1}C_n$ is, by binomial theorem:

$$\sum_{K=0}^{K=n}e^{2 i \pi KL/n}\ \binom{K}{n}=(1+e^{2 i \pi L/n})^n$$

$$=e^{i \pi L} \ (e^{-i \pi L/n}+e^{i \pi L/n})^n$$

whose modulus is a $2^n \cos(\pi L/n)^n \ \ (0 \leq L \leq n)$.

which is (up to a shift) reasonably close to the coefficient of $C_n$. Why that ?

A continuous analog will help in the understanding of this point:

Take the approximation $\cos^n(x) \approx (1-\frac{x^2}{2})^n$, then renormalize it (taking into account a certain standard deviation) into $(1-\frac{(x/\sqrt{n})^2}{2})^n$ $= (1-\frac{x^2}{2n})^n$ which tends to $e^{-x^2/2}$ when $n$ tends to $+\infty$ (classical limit), a quite satisfactory result.