Normal distribution pdf expressed through cosine

Question

I have had multiple attempts at squeezing a normal distribution into a finite domain for like last five years. I find this important as sometimes random value should be limited to certain real boundaries while preserving the shape of normal distribution and it doesn't feel comfortable just to cut both tails of normal distribution and say that's fine.

While attempting to do so, I came up with this interesting (at least to me) formula:

$$\lim_{n \to \infty} \frac{\left[1 + \cos\left[x\sqrt{2}\frac{\Gamma\left[\frac{1}{2}+n\right]}{\Gamma\left[1+n\right]}\right]\right]^n}{2^{\frac{1}{2}+n}\sqrt{\pi}}=\frac{\exp{\left[-\frac{x^2}{2}\right]}}{\sqrt{2\pi}}$$

Which is the probability density function of a normal distribution $$N(0,1)$$

I had never encountered this specific representation of a normal distribution before. Clearly, Euler's formula defines a link between trigonometric functions and exponents:

$$e^{ix}=\cos(x)+i\sin(x).$$

This specific representation makes so much sense so I'm curious if this representation was long known before.

Answer 1

The upshot is that this isn't a really interesting connection with $\cos$ since the argument of the cosine converges to $0$, and so $\cos$ can be replaced with any function $f$ such that $f(x) = 1 - x^2/2 + O(x^3)$ as $x\rightarrow 0$.

More precisely, we have that $$ \frac{\Gamma(1/2 + n)}{\Gamma(1 + n)} = n^{-1/2} + O(n^{-3/2}), $$ which allows us to write $$ \cos\biggl(x\sqrt{2} \frac{\Gamma(1/2 + n)}{\Gamma(1 + n)}\biggr) = \cos\biggl(x\sqrt{\frac{2}{n}} + O(n^{-3/2})\biggr) = 1 - \frac{x^2}{n} + O(n^{-3/2}), $$ from which the result follows.

But we could have made the same argument with, say $f(x) = 2 - \cosh(x)$, from which would follow the identity $$ \lim_{n\rightarrow\infty} \biggl[3 - \cosh\biggl(x\sqrt{2}\frac{\Gamma(1/2 + n)}{\Gamma(1 + n)}\biggr)\biggr]^n/2^n = e^{-x^2/2} $$ or even $f(x) = \sqrt{1 - x^2}$, rendering $$ \lim_{n\rightarrow\infty} \biggl[1 + \sqrt{1 - \biggl(x\sqrt{2}\frac{\Gamma(1/2 + n)}{\Gamma(1 + n)}\biggr)^2}\biggr]^n/2^n = e^{-x^2/2}. $$