Apologies in advance for the length... I tried to be concise but it's hard...
Whenever people talk about how $e$ and $\pi$ are connected, they always talk about Euler's identity $(e^{i\pi}=-1)$, and while this is fascinating and beautiful and all of that, it's hard to visualize or conceptualize this connection because it involves the very unintuitive complex number $i$, which makes the connection to reality a bit dubious (at least to me).
Compare this to the equation of the normal curve, which I'm sure you all know is: $${\phi(Z)={1\over\sqrt{2\pi}}e^{-{1\over2}Z^2}},$$ where $Z$ is a random variable with a standard normal $(0,1)$ distribution.
Here we have an equation that has a fairly obvious connection to reality (explained by the central limit theorem), which brings $e$ and $\pi$ together, and I'm wondering if there is some explanation for why this is. Surely, it has to be more than a coincidence.
I already understand why the coefficient $1\over{\sqrt{2\pi}}$ is added to the equation for the standard normal curve, but this does not answer the question of what the connection is, since:
$$f(x)=e^{-{1\over2}x^2} \Rightarrow \int_{-\infty}^\infty{f(x)dx}=\sqrt{2\pi},$$
and obviously $f(x)$ has the exact same fundamental shape, with all the same fundamental characteristics (except for area being 1) as the standard normal curve and is only being scaled by $1\over{\sqrt{2\pi}}$ to make the area under the curve more suitable to calculating probabilities.
My point here is that regardless of whether you consider $f(x)$ or $\phi(x)$, $\pi$ is lurking right beside $e$ and I would love to know what is going on here.
I'm a fan of 3blue1brown on youtube, and he insists that whenever $\pi$ is lurking somewhere, there is some connection to a circle, but I'm not seeing it.
So my first question here is, how exactly does the normal curve relate to a circle?
My understanding of the central limit theorem is that if you were to draw a histogram of infinitely many sums or averages of random variables with the same distribution, as the number of sums or averages being considered approaches infinity, the shape produced by the histogram will begin to perfectly line up with a scaled version of the normal curve.
Assuming I've got this right (perhaps someone could confirm this for me and correct me if I'm mistaken), this means that both $\pi$ and $e$ and lurking every time we take the sum or average of similarly distributed things, which is something we all do quite frequently.
My main question here is this: what exactly is going on here?
The fact that $e$ and $\pi$ are side-by-side whenever you sum or average something can't be a coincidence, and every time I try to find out what's going on, my search results are drowned by people discussing Euler's identity.
I would appreciate it very much if someone could enlighten me, or direct me somewhere that has the answers I'm looking for.
There is indeed a connection between the circle and the equation $\int_{-\infty}^{\infty} e^{-x^2/2} \, dx = \sqrt{2\pi}$. The connection comes from the proof of that equation. Namely, \begin{align*} \left(\int_{x \in \mathbb{R}} e^{-x^2/2} \, dx\right)^2 &= \left(\int_{x \in \mathbb{R}} e^{-x^2/2} \, dx\right) \cdot \left(\int_{x \in \mathbb{R}} e^{-x^2/2} \, dx\right) \\ &= \left(\int_{x \in \mathbb{R}} e^{-x^2/2} \, dx\right) \cdot \left(\int_{y \in \mathbb{R}} e^{-y^2/2} \, dy\right)\\ &= \int\!\!\!\int_{(x,y) \in \mathbb{R}^2} e^{-x^2/2} \cdot e^{-y^2/2} \, dx \, dy \\ &= \int\!\!\!\int_{(x,y) \in \mathbb{R}^2} e^{-(x^2+y^2)/2} \, dx \, dy \\ &= \int\!\!\!\int_{r > 0, \,\, 0 \le \theta \le 2\pi} e^{-r^2/2} r \, dr \, d\theta \end{align*} The last line is where circles have entered the picture, by changing variables from Euclidean coordinates $x,y$ to polar coordinates $r,\theta$. That last integral is easy to evaluate, and its value comes out to be $2\pi$.