Perimeter of an ellipse How can I calculate the perimeter of an ellipse? What is the general method of finding out the perimeter of any closed curve?

If the semi-major axis has length $a$ and the eccentricity is $e$, then the perimeter is
$$
2πa \left(\frac{1}{1} \left(\frac{(-1)!!}{0!!} e^0\right)^2 + \frac{1}{-1} \left(\frac{1!!}{2!!} e^1\right)^2 + \frac{1}{-3} \left(\frac{3!!}{4!!} e^2\right)^2 + \frac{1}{-5} \left(\frac{5!!}{6!!} e^3\right)^2 + ⋯\right)
$$
where $(-1)!! = 1 = 0!!$, and for all $n ∈ \{⋯, -7, -5, -3, -1, 0, 1, 2, 3, 4, 5, ⋯ \}$: $(n+2)!! = (n+2) n!!$. The sum extends to negative powers of $e$, if you take $n!! = ∞$ for $n ∈ \{⋯, -8, -6, -4, -2\}$; and if you work out the double-factorials, using:

*

*$(2n)!! = 2^n n!$, for $n ≥ 0$;

*$(2n-1)!! = \frac{(2n)!}{2^n n!}$, for $n ≥ 0$;

*$(2n-1)!! = \frac{(-1)^n}{(-2n - 1)!!} = (-2)^{-n} \frac{(-n)!}{(-2n)!}$, for $n ≤ 0$;

then it comes out to the following:
$$
2πa \left(1 - \left(\frac{1}{2} e\right)^2 - \frac{1}{3} \left(\frac{1}{2} \frac{3}{4} e^2\right)^2 - \frac{1}{5} \left(\frac{1}{2} \frac{3}{4} \frac{5}{6} e^3\right)^2 + ⋯\right).
$$
Im my previous life oops I said that out loud I offered the following estimate:
$$\mbox{Ramanujan's Formula}:
π(3(a+b) - \sqrt{(3a+b)(a+3b)}) = π(a+b) (3 - \sqrt{4 - h}),$$
where $b = a \sqrt{1 - e^2}$ is the length of the semi-minor axis and $h = ((a - b)/(a + b))^2$, while secretly holding onto the other, much better, estimate:
$$π(a+b) \left(\frac{12 + h}{8} - \sqrt{\frac{2 - h}{8}}\right).$$
