Perimeter of an ellipse

Unlike most other elementary shapes, such as the circle and square, there is no closed-form expression for the perimeter of an ellipse. Throughout history, a large number of closed-form approximations and of expressions in terms of integrals or series have been given for the 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

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

where ${\displaystyle e}$is the eccentricity of the ellipse, defined as

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

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

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

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, Bessel and Kummer, 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 terns 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.

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

Indian mathematician Srinivasa Ramanujan proposed multiple approximations.

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

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

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

$${\displaystyle \varepsilon \approx {\dfrac {3ae^{20}}{68719476736}}}$$

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

Ramanujan did not provide any rationale for this formula.

$${\displaystyle P\approx 2\pi {\sqrt {\frac {a^{2}+b^{2}}{2}}}.}$$

This formula is simpler than most perimeter formulas but less accurate for highly eccentric ellipses.

In more recent years, computer programs have been used to find and calculate more precise approximations of the perimeter of an ellipse. In an online video about the perimeter of an ellipse, recreational mathematician and YouTuber Matt Parker, using a computer program, calculated numerous approximations for the perimeter of an ellipse. Approximations Parker found include

$${\displaystyle P\approx \pi \left({\frac {6a}{5}}+{\frac {3b}{4}}\right)}$$

and

$${\displaystyle P\approx \pi \left({\frac {53a}{3}}+{\frac {717b}{35}}-{\sqrt {269a^{2}+667ab+371b^{2}}}\right).}$$

• Meridian arc#Full meridian
• Ellipse#Circumference