Perimeter of Ellipse

[Albert Chan]

From previous 2 posts, we can get some nice Elliptic Integral Identity:
Below, all Elliptic functions uses modulus argument.

Eccentricity of ellipse, e^2 = 1 - (b/a)^2
We proved "next" ellipse (a' = (a+b)/2 , b' = sqrt(a*b)), has eccentrity of h = (a-b)/(a+b)

K(e) = pi/2 / agm(1,b/a) = pi/2 / agm(1+e,1-e)
K(h) = pi/2 / agm(1+h,1-h) = pi/2 / agm(1,b/a) / (1+h)

→ K(e) = (1+h) K(h)

(01-21-2020 05:16 PM)Albert Chan Wrote:  \(\large p(a,b) = 2× \left( p({a+b \over 2}, \sqrt{ab}) - {\pi a b \over AGM(a,b)}\right)\)

With ellipse_perimeter = 4 a E(e), we have:

4 a E(e) = 4 (a+b) E(h) - 4 b K(e) = 4 (a+b) E(h) - 4 b (1+h) K(h)

→ E(e) = 2/(h+1) * E(h) - (1-h) * K(h)