Why is $π$ ubiquitous?

Question

It is not confusing to find $π$ in the area of the circle equation or its circumference, But when I started studying math and physics a little bit deeper, I begin to see $\pi$ in very strange positions for instance: $\sum_{n=1}^∞ \frac {1}{n^2} = \frac{\pi^2}{6}$, Or in Coulomb's constant $k=\frac{1}{4\pi\epsilon}$, Or in Euler's identity $e^{i\pi }+1=0$ and in many other places.

I am not asking about these specific positions, I am asking generally why is $\pi$ everywhere and confusion positions ?

Answer 1

For $\sum_{n=1}^\infty \frac 1 {n^2} = \frac{\pi^2}6$ there is a very nice video explanation by 3Blue1Brown available at YouTube, relating the sum to circles and hence to $\pi$.

For Euler's identity note that $e^{ix} = \cos(x)+i\sin(x)$ relates exponentials to circles immediately and then $e^{i\pi}=-1$ is merely saying "walking 180° around the unit circle starting at $1$, you end up at $-1$".

I'm not sure about $\pi$s appearance in Coulumb's constant as I'm not a physicist, but I'm guessing circles (or more generally spheres) play a role here too.

Answer 2

As you noted, $\pi$ is related to circles. Therefore it is also related to angles, or surfaces. A very general way into which "angles" appear are Fourier series, which are relevant for any periodic phenomenon (and more).

The easiest way (that I know of) to compute $\sum_{n=1}^\infty \frac{1}{n^2}$ is via Fourier series.

Again, in the formula $e^{i\pi}+1=0$, $\pi$ plays the role of an angle.

In the case of Coulomb's constant, it appears because you integrate over the surface of a sphere.

Answer 3

A friend of mine said "I used to think of pi as related to circles; this article set me straight." and pointed me to

https://affinemess.quora.com/What-is-math-pi-math-and-while-were-at-it-whats-math-e-math

which I highly recommend.

Answer 4

About Coulomb's constant:
Gauss's law states that $$\operatorname{div}(E)=\frac{\varrho}{\varepsilon_0}$$ Or alternatively (the integral form): $$\int E \cdot \hat{n} \mathrm{d}S = \frac{q}{\varepsilon_0}$$ And if we assume that there's a point charge at $0$ and spherical symmetry, then we have that $$E(r)=|E(r)| \hat{n}$$ So if we integrate it over a sphere centered at $0$ with radius $r$ we get that $$|E(r)| 4 \pi r^2=\frac{q}{\varepsilon_0}$$ $$|E(r)|=\frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2}$$ And it's convenient to let $$k=\frac{1}{4 \pi \varepsilon_0}$$

Answer 5

$2i\pi$ is the period of the complex exponential. The inverse of this function is the logarithm, known to have the derivative $\frac1z$. That creates a close connection to the poles of complex functions, which makes $2i\pi$ appear in the residue formula, and corresponds to a phase jump; in geometric terms, a full turn.

In connection with the polar coordinates, it appears in several integrals with circular symmetry.

• the perimeter of the circle,

• the area of the circle,

• the volume under a bivariate Gaussian surface.

The latter integral is related to the Gamma function and explains why $\Gamma(\frac12)=\sqrt\pi$. I guess that this is also related to the $\sqrt\pi$ that appears in the Stirling formula for the factorial.

By a factorization process (a consequence of Fubini's theorem ?), $\pi$ ends-up in integrals that generalize those of the circle (area and volume of hyperspheres), with $\pi$ to powers depending on the dimensions (and involving $\Gamma$ of half-integers). This is why you find $\pi$ in some physics formulas having to do with spherical symmetry.

The Basel problem (sum of inverses of squares) can be proven from a factorization of the sine, itself related to the period. The generalization to higher degrees (sum of inverses of even powers) also leads to higher powers of $\pi$, this time together with the Bernouilli numbers.