Why is the ratio of circumference to diameter $\pi$, and not any other number?

Question

How do we specifically get to 3.14 from axioms through logic to that particular transcendental number being the case?

I'm guessing it's the sum of an infinite series. But how do we go from "there's a circle" to "this particular infinite series describes the ratio of the diameter to the circumference", other than getting out a tape measure, and noticing that the series seems to produce the same number (sans any proof). I'm looking for the proof of why the circumference/diameter = 3.14... specifically.

Answer 1

Consider the circle of radius $R$ centered at the origin. That is, the set of points such that $x^2+y^2=R^2$. The upper half of the circle is described by $y=f(x)=\sqrt{R^2-x^2}$. So if we call $\mu(R)$ the circle's perimeter, we have (from the arc-length formula) \begin{align} \mu(R)&=2\int_{-R}^R\sqrt{1+f'(x)^2}\,dx=2\int_{-R}^R\sqrt{1+\left(-\frac{x}{\sqrt{R^2-x^2}}\right)^2}\,dx\\ \ \\ &2=\int_{-R}^R\frac{R}{\sqrt{R^2-x^2}}\,dx=2\int_{-R}^R\frac1{1-(x/R)^2}\,dx\\ \ \\ &=2R\,\int_{-1}^1\frac1{\sqrt{1-x^2}}\,dx. \end{align} This in particular tells us that the perimeter is proportional to the radius. If we want to talk in terms of the diameter, we get $$ \mu(D)=D\,\int_{-1}^1\frac1{\sqrt{1-x^2}}\,dx. $$ If follows that the quotient $\mu(D)/D$ is a constant, that we name $\pi$, and is equal to $$ \pi=\int_{-1}^1\frac1{\sqrt{1-x^2}}\,dx. $$ If you didn't know anything else about $\pi$, you could now use approximations of the integral to calculate approximations of $\pi$.

For better convergence of the Riemann sums, one can easily show (via integration by parts) that $$ \int_{-1}^1\frac1{\sqrt{1-x^2}}\,dx=2\int_{-1}^1{\sqrt{1-x^2}}\,dx $$ (which in particular also shows that $\pi$ is the area of the unit disk)

Answer 2

This is a very abbreviated development from Rudin's "Real & Complex Analysis".

Start with the Taylor series for $\exp$. Show that $t \mapsto \exp(it)$ maps the reals onto the unit circle.

Define $\sin, \cos$ as the imaginary & real parts of $\exp$. Show $\cos'= -\sin, \sin'=\cos$. We have $\cos 0 = 1$ and with a little work can show there is a smallest positive number $t_0$ for which $\cos t_0 = 0$. Define $\pi= 2 t_0$.

Answer 3

Using the Archimedes approach. Start with n equally spaced radii, each of length $\frac{1}{2}$. For each pair of radii form a triangle, where the third side is the chord connecting the end points of the radii. The sum of the chord lengths (which approximates the length of the circumference) will $\to \pi$, as n$\to \infty$.