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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 $\pi$?

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  • $\begingroup$ 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

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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

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

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  • $\begingroup$ Why do you write $\text{Pi}$ rather than $\pi\text{?} \qquad$ $\endgroup$ Mar 3, 2020 at 15:52
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    $\begingroup$ @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
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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.

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

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

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