Perimeter of an ellipse

An ellipse is defined by two axes: the major axis (the longest diameter) of length ${\displaystyle 2a}$ and the minor axis (the shortest diameter) of length ${\displaystyle 2b}$, where the quantities ${\displaystyle a}$ and ${\displaystyle b}$ are the lengths of the semi-major and semi-minor axes respectively. The exact perimeter ${\displaystyle P}$ of an ellipse is given by the integral^[1]

$${\displaystyle P=4a\int _{0}^{\pi /2}{\sqrt {1-e^{2}\sin ^{2}\theta }}\ d\theta ,}$$

where ${\displaystyle e}$ is the eccentricity of the ellipse, defined as^[2]

$${\displaystyle e={\sqrt {1-{\frac {b^{2}}{a^{2}}}}}.}$$

If we define the function

$${\displaystyle E(x)=\int _{0}^{\pi /2}{\sqrt {1-x\sin ^{2}\theta }}\ d\theta ,}$$

known as the complete elliptic integral of the second kind, the perimeter can be expressed in terms of that function as simply

$${\displaystyle P=4aE(e^{2}).}$$

The integral used to find the perimeter does not have a closed-form solution in terms of elementary functions.

Another solution for the perimeter, this time using the sum of a infinite series, is^[3]

$${\displaystyle P=2a\pi \left(1-\sum _{n=1}^{\infty }{\frac {(2n!)^{2}}{(2^{n}\cdot n!)^{4}}}\cdot {\frac {e^{2n}}{2n-1}}\right),}$$

where ${\displaystyle e}$ is the eccentricity of the ellipse.

More rapid convergence may be obtained by expanding in terms of ${\displaystyle h=(a-b)^{2}/(a+b)^{2}}$. Found by James Ivory,^[4] Bessel^[5] and Kummer,^[6] there are several equivalent ways to write it. The most concise is in terms of the binomial coefficient with ${\displaystyle n=1/2}$, but it may also be written in terms of the double factorial or integer binomial coefficients: $${\displaystyle {\begin{aligned}{\frac {P}{\pi (a+b)}}&=\sum _{n=0}^{\infty }{1/2 \choose n}^{2}h^{n}\\&=\sum _{n=0}^{\infty }\left({\frac {(2n-3)!!}{(2n)!!}}\right)^{2}h^{n}\\&=\sum _{n=0}^{\infty }\left({\frac {(2n-3)!!}{2^{n}n!}}\right)^{2}h^{n}\\&=\sum _{n=0}^{\infty }\left({\frac {1}{(2n-1)4^{n}}}{\binom {2n}{n}}\right)^{2}h^{n}\\&=1+{\frac {h}{4}}+{\frac {h^{2}}{64}}+{\frac {h^{3}}{256}}+{\frac {25\,h^{4}}{16384}}+{\frac {49\,h^{5}}{65536}}+{\frac {441\,h^{6}}{2^{20}}}+{\frac {1089\,h^{7}}{2^{22}}}+\cdots .\end{aligned}}}$$ The coefficients are slightly smaller (by a factor of ${\displaystyle 2n-1}$) than the preceding, but also ${\displaystyle e^{4}/16\leq h\leq e^{4}}$ is numerically much smaller than ${\displaystyle e^{2}}$ except at ${\displaystyle h=e=0}$ and ${\displaystyle h=e=1}$. For eccentricities less than 0.5 (${\displaystyle h<0.005}$), the error is at the limits of double-precision floating-point after the ${\displaystyle h^{4}}$ term.^[7]

Because the exact computation involves elliptic integrals, several approximations have been developed over time.

Indian mathematician Srinivasa Ramanujan proposed multiple approximations.^[8][9]

$${\displaystyle P\approx \pi \left(3(a+b)-{\sqrt {(3a+b)(a+3b)}}\right).}$$

$${\displaystyle P\approx \pi (a+b)\left(1+{\frac {3h}{10+{\sqrt {4-3h}}}}\right),}$$

where ${\displaystyle h={\frac {(a-b)^{2}}{(a+b)^{2}}}}$.

The final approximation in Ramanujan's notes on the perimeter of the ellipse is regarded as one of his most mysterious equations. It is

$${\displaystyle P=\pi \left((a+b)+{\frac {3(a-b)^{2}}{10(a+b)+{\sqrt {a^{2}+14ab+b^{2}}}}}+\varepsilon \right),}$$

where

$${\displaystyle \varepsilon \approx {\dfrac {3ae^{20}}{2^{36}}}}$$

and ${\displaystyle e}$ is the eccentricity of the ellipse.^[9]

Ramanujan did not provide any rationale for this formula.

• Meridian arc#Full meridian
• Ellipse#Circumference