Deriving $\pi$ from $x^2+y^2=1$

Question

When I was on the road today I came across one interesting puzzle.

What if you meet an extraterrestrial species, and you show to them some of mathematical works. They look at $\pi$ and they clearly don't understand what it means. So they ask you, and you write $x^2 + y^2 = 1$ and say, "Look, $\pi$ comes from it."

How to show the exact details? What way will be shorter and why?

I came up with 3 basic ways but none of them is easy, each of them involves some geometry. But universally there have to be built some trigonometry around, otherwise at integration/differentiation there would be misunderstanding of arcsin/arctan functions. Also $\lim\frac{\sin x}{x}$ have to be built somehow, and geometry is best suited for it.

Is there any way to do it without geometry at all? What is the shortest path from $x^2+y^2=1$ to $\pi$?

Answer 1

We can compute the length of the circumference of a unit half-circle with

$$x^2+y^2=1\implies 2x+2yy'=0$$ so that $$\text{Pi}=\int_{-1}^1\sqrt{1+y'^2} \, dx = \int_{-1}^1 \frac{dx}{\sqrt{1-x^2}}.$$

Then by Taylor,

$$\frac1{\sqrt{1-x^2}}=\sum_{k=0}^\infty\frac{(\frac12-k)!}{k!} x^{2k}$$

and

$$\text{Pi}=2\sum_{k=0}^\infty\frac{(\frac12-k)!}{(2k+1)k!}.$$

Answer 2

I have come to really marvellous high-school tier single-step solution which routes to the same Leibnitz series, but involves only one step with both geometry and vector multiplication.

Below is typeset

$$ \mathbf{e} \times \mathbf{de} = \sin(\hat{e;de})\cdot|e|\cdot|de| $$ $$ dl = |de| = \frac{1}{|e|} \mathbf{e}\times\mathbf{de}=\frac{x\,dy-y\,dx}{x^2+y^2} $$ $$ \int_0^{L/4}dl=\int\frac{1}{1+\left(\frac{y}{x}\right)^2}\frac{x\,dy-y\,dx}{x^2}=\int_0^1 \frac{d(y/x)}{1+(y/x)^2} $$

Below is classic Leibnitz solution (3 steps).

1. Consider $\triangle OAB \propto \triangle AKM $ so for sides when we obtain infinitesimally small $\triangle AKM$, we got $$\frac{-dx}{y}=\frac{dy}{x}=\frac{dl}{1}=\frac{d\phi}{\kappa},$$ here we acknowledge that angles are proportional to arc with coefficient $\kappa$ because we able to stack many equal triangles together. I tried to exclude this geometry step somehow but it seems extending the whole thing.

2. Now we define our ratio for alien (tangent func) as $t=\frac{y}{x}$. From (1) and next with integration it follows $$\frac{dx}{d\phi}=\frac{-y}{\kappa},\frac{dy}{d\phi}=\frac{x}{\kappa}.$$ $$\frac{dt}{d\phi}=\frac{y'x-yx'}{x^2}=\frac{x^2+y^2}{\kappa x^2}$$ $$\frac{d\phi}{dt}=\kappa x^2=\frac{\kappa}{1+t^2}$$

3. Now we denote straight angle $\phi_0$, arc $L$, we know that $t=1$ at one forth of it from symmetry considerations. Here we get Leibnitz: $$\int_0^{L/4} dl = \int_0^{\phi_0/4} \frac{d\phi}{\kappa} = \int_0^1 \frac{dt}{1+t^2}.$$

$$\pi=L=\frac{\phi_0}{\kappa}=4 \sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1} $$