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 $\pi$?

  • $\begingroup$ In regards to your (previous) title: $\pi$ is transcendental, not algebraic $\endgroup$ Mar 3, 2020 at 14:52
  • $\begingroup$ Yea, I know I used bad wording $\endgroup$
    – sanaris
    Mar 3, 2020 at 14:55
  • $\begingroup$ First of all, you should define $\pi$ in some way. If you define it as the "length of the circunference", you should probably need something like a limit or the integral of a derivative to understand what the "length of a curve" is. Same questioning may happen if you define it using areas. Tell me what definition of $\pi$ you are using and what is within the scope of "admisible" steps when reading $\pi$ off that circunference equation. $\endgroup$ Mar 3, 2020 at 14:57
  • $\begingroup$ To give you an example: if you like the arc length stuff, you can write $y=\sqrt{1-x^2}$ from your equation. That defines a function on $[-1,1]$ whose arc length is exactly $\pi$. Hence if you call $f(x)=\sqrt{1-x^2}$ you just have: $$\pi = \int_{-1}^1 \sqrt{1+f'(x)^2} dx$$ $\endgroup$ Mar 3, 2020 at 15:06
  • $\begingroup$ For other interesting and unexpected applications of $\pi$ see this question: math.stackexchange.com/questions/689315/… $\endgroup$ Mar 3, 2020 at 15:16

2 Answers 2


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,

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



  • $\begingroup$ Why do you write $\text{Pi}$ rather than $\pi\text{?} \qquad$ $\endgroup$ Mar 3, 2020 at 15:52
  • 1
    $\begingroup$ @MichaelHardy: just in case this definition would not coincide with the known one, due to… hem… quantum effects. ;-) $\endgroup$
    – user65203
    Mar 16, 2020 at 7:48

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.

enter image description here

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). Leibnitz tangent solution

  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: $$\int_0^{L/4} dl = \int_0^{\phi_0/4} \frac{d\phi}{\kappa} = \int_0^1 \frac{dt}{1+t^2}.$$

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.