What exactly ARE $\pi$ and $e$?

Question

First of all, apologies if this is a bad question. I don't really know how to phrase it.

I first got introduced to $\pi$ in elementary school, where it was presented as a ratio for a circle's area. I thought it was some special number that had to do with circles. You can imagine my confusion when I found it popping up in stuff that seemingly had nothing to do with circles, like in the Wallis formula or the Basel Problem. Same with $e$. I thought it was this financial growth thing, so I got pretty confused when it showed up in stuff like the Probability Distribution, and Euler's identity. Right now, they seem like two magic numbers with magical properties - same thing with $\cos$ and $\sin$ - magical equations that give you ratios of lengths.

Could someone explain what the heck exactly is the significance of $e$ and $\pi$? Beyond circles, beyond geometry. Because they certainly seem to be more than just that.

Answer 1

I would recommend you read the article on wikipedia about the definition of $\pi$. It is suggested to start from the cosine function, which can be defined as a series or a differential equation. The reason is that usually in calculus you learn series and derivatives before integral. If you have this notions, then one can derive the circumference of the circle by integration, so $\pi$ as defined in antiquity is a consequence of a calculation performed 3000 years later. Similar with $e$, if you start with either the Taylor series representation or the basic definition that $\frac{de^x}{dx}=e^x$, you can then derive all properties you described in your question.

Answer 2

This may surprise you, but every appearance of $\pi$, no matter how crazy-looking it is, traces to the involvement of a circle somewhere.

For example, where does the Wallis product come from? From evaluating $\frac{\sin x}{x}=\prod_{k\ge 1}(1-\frac{x^2}{k^2})$ at $x=\frac{\pi}{2}$ (i.e. a quarter-turn from $x=0$). And though we teach trigonometry to students with right-angled triangles, you can embed those triangles in a circle of which the hypotenuse is a radius, and generalise the cosine and sine beyond acute angles so that, if in the circle $x^2+y^2=1$ we consider a radius $(1,\,0)$ and another forming an anticlockwise angle $\theta$ with it, the latter meets the circle at $(\cos\theta,\,\sin\theta)$. So ultimately, these functions are defined with circles.

What about the Basel problem? We can prove $\sum_{k\ge 1}k^{-2}=\frac{\pi^2}{6}$ by using the fact that acute $x$ satisfy $\cot^2x\le x^{-2}\le\csc^2 x=1+\cot^2 x$, which we sum over $x\in\{\frac{k\pi}{2m+1}|1\le k\le m\}$. (The sum of the squared cotangents can be obtained from a polynomial's coefficients.) So once again, it leads back to trigonometry, hence circles.

Similarly, every appearance of $e$, no matter how crazy-looking it is, traces to either of its equivalent definitions (1) $e:=\lim_{n\to\infty}(1+\frac{1}{n})^n$ or (2) $\frac{d}{dx}e^x=e^x,\,e>0$. (1) implies $(1+\frac{x}{n})^n\approx e^x$ for $x\ll n$, from which we can deduce the central limit theorem using characteristic functions. But this proof also uses $e^{ix}=\cos x+i\sin x$, which can be proven by combining (2) with a proof that $\cos x,\,\sin x$ satisfy $y''=-y$.

Why does $\pi$ keep coming up? Because multivariable symmetries frequently concern tracing over the surface of a circle, sphere or hypersphere, which traces back to circles. Why does $e$ keep coming up? Because so many problems hinge on rates of change, which automatically include exponential functions if the differential equations involved take a certain form.