I am studying trigonometry and I know that $\pi$ can be approximated using Gregory series, Rutherford series, etc. Also it's strange and mysterious that $\pi$ is just ratio of circumference to diameter. This profoundly shows integration in math between different branches.
I would like to know other ways of $\pi$ approximation in different fields of math. Please share your views.
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The number of reduced fractions $p/q$ with $1\le p\le q\le n$ is approximately $(3/\pi^2)n^2$ --- approximately, in the sense of asymptotically equal (the ratio of the two quantities approaches 1 as $n\to\infty$). So $\pi$ can be approximated by counting the number of reduced fractions $p/q$ with $1\le p\le q\le n$, dividing by $n^2$, taking the reciprocal, multiplying by 3, and taking the square root. For example, for $n=7$, the reduced fractions are $1/1,1/2,1/3,2/3,1/4,3/4,1/5,2/5,3/5,4/5,1/6,5/6,1/7,2/7,3/7,4/7,5/7,6/7$, 18 of them; what approximation does that give you for $\pi$?
Also, the number of squarefree naturals up to $n$ (meaning, naturals not divisible by the square of any prime) is asymptotically $(6/\pi^2)n^2$. So $\pi$ can be approximated by counting the number of squarefree naturals up to $n$, dividing by $n^2$, taking the reciprocal, multiplying by 6, and taking the square root. For example, for $n=30$, the squarefree numbers are $1,2,3,5,6,7,10,11,13,14,15,17,19,21,22,23,26,29,30$, 19 of them. What approximation do you get for $\pi$?
On
$\pi/2$ is the least positive solution of $0=f(x)=(\sin x) -1.$ Since $f'(x)>0 >f(x) $ for $x\in (0,\pi/2)$ we can use Newton's method to approximate $\pi/2:$ Take any $x_1\in (0,\pi/2)$ and let $$x_{n+1}=x_n-f(x_n)/f'(x_n). $$ We can use the angle-sum formulas together with the power series for $\sin$ and $\cos, $ to more easily compute $\sin x_{n+1}$ and $\cos x_{n+1}$ from $\sin x_n$ and $\cos x_n.$ Of course $x_n\to \pi/2$ as $n\to \infty.$
Since $f'(\pi/2)=0,$ a faster convergence is obtained if we let $x_{n+1}=x_n-2f(x_n)/f'(x_n).$
There are many variations on this theme.
On
The roots of $\frac{\sin{x}}{x}$ are the non-zero roots of its numerator $\sin{x}$: $\pm\pi$, $\pm 2\pi$, $\pm 3\pi$... and so on.
Using these roots, Euler factored the series for $\frac{\sin{x}}{x}$ as the infinite product: $$\frac{\sin{x}}{x}=\prod_{n=1}^\infty \left(1-\frac{x^2}{n^2\pi^2}\right)$$ If you use Euler's formula and set $x=\frac{\pi}{2}$, you can easily get $$\pi=\frac{2}{\prod_{n=1}^\infty \left(1-\frac{1}{4 n^2}\right)}.$$
The half circumference of the unit circle can be computed from the implicit equation $x^2+y^2=1$, which expresses a constant distance to the origin.
$$H=\int_{-1}^1\sqrt{y'^2(x)+1}\ dx=\int_{-1}^1\frac{dx}{\sqrt{1-x^2}}=\arcsin x\Big|_{-1}^1=\pi.$$
Similarly, for the area. $$A=2\int_{-1}^1y\ dx=2\int_{-1}^1\sqrt{1-x^2}\ dx=\left(x\sqrt{1-x^2}-\arcsin x\right)\Big|_{-1}^1=\pi.$$ When you accept the analytic formula for the Euclidean distance, you make a connection between geometry and calculus. And from there the link with trigonometric functions and the power series that allow to compute the value of $\pi$.
By the way, numerical estimates of these integrals provide approximations for $\pi$.