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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 ?

Jam
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Mohammad Alshareef
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  • Because $\pi$ is an important mathematical constant? Because circle is a very important concept that is central to trigonometry? Because quite a lot of our mathematics arose from planar geometry which was important from early Greece to 18 century France? I don't know, you are free to choose your own reason. – Vasily Mitch May 14 '19 at 10:49
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    Basically anything that involves any kind of circle or sphere, is going to involve pi as well – Rage May 14 '19 at 10:53
  • There as many sources on the web, for example https://en.wikipedia.org/wiki/Pi#Use_in_mathematics_and_science – Jean Marie May 14 '19 at 10:55
  • The appearance of π in the Coulomb formula is related to circles. The $4πr^2$ in the denominator is because we are finding the flow of the electric field per unit of area through the surface of a sphere, whose surface area is $4πr^2$. – MJD May 14 '19 at 12:04
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    The $e^{iπ}=-1$ one is also circle-related. It asks how far around a unit circle you have to go, in order to be travelling in the opposite direction ($-1$) from when you started; the answer is that you must travel a distance of $π$. (Or $3π$, but $e^{3πi}=-1$ as well.) Note similarly that $e^{2πi}=+1$. – MJD May 14 '19 at 12:14
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    @MJD: the fact that the same $\pi$ appears both in the formula for the perimeter and area of a circle it in itself an intriguing coincidence. But That it also appears in the surface and the volume of a sphere is a bigger surprise. –  May 14 '19 at 13:17
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    Once you know that π is involved in the perimeter and area of a circle, the fact that it also appears in the surface and volume of a sphere should not be surprising. It would be much more surprising, in my opinion, if it didn't also appear there. (“Where did the π go?”) – MJD May 14 '19 at 15:25
  • @mohammed I should have emphasized that the appearance of π in the Coulomb constant is really just a choice of units. It appears in the SI value of the Coulomb constant because of the way the units work out (and because as I said, in SI we are calculating electric flux per unit area) but if you use the Gaussian units the value of the constant is precisely 1, no π involved. Nothing has changed except the size of the units, analogous to the way meters per second are exactly the same as miles per hour but with different constants. – MJD May 14 '19 at 15:30
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    While I think that this is an interesting question, I also think that it fits poorly with the MSE format. It is not terribly specific and not really amenable to a single answer. It is also problematic in that it is, perhaps, more about the philosophy of mathematics, rather than mathematics itself. As such, I am voting to close. – Xander Henderson May 15 '19 at 13:34

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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.

Christoph
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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.

John Hughes
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  • interesting points, but I would argue that the complex exponential function IS related to angles... – Albert May 14 '19 at 13:33
  • Sure. And electricity is related to amber. :) Seriously, though: the point isn't that $\pi$ is unrelated to angles --- it's that it occurs in the solution to the most basic of nontrivial ODEs, and that helps explain its occurrence elsewhere. Still, if you don't find that particular view of things enlightening, and prefer to see everything through angles (or amber!), that can work too. Different strokes, etc. – John Hughes May 14 '19 at 14:21
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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.

Albert
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$2i\pi$ 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.

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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 $$k=\frac{1}{4 \pi \varepsilon_0}$$

Botond
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