$Deriving Ramanujan's Ellipse approximation$

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$Deriving Ramanujan's Ellipse approximation$

Ramanujan's ellipse approximation.

Taking the major and minor axises of an ellipse to be a and b, first you compute;

\[h = (a – b)/(a + b).\]

Then his approximation is the following.

\[\pi (a + b) \left(1 + \frac{3h^2}{10 + \sqrt{4 - 3h^2}}\right)\]

I can see how π•(a+b) can make sense as the closer a is to b the more circular the ellipse and the more accurate the approximation is as you basically have π•radius+radius or π•diameter. I do not understand how he got the rest of his equation. If anyone can show me how it's derived it'd be greatly appreciated!

Geometry

Frontblunt333

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See the Gauss-Kummer formula in section 7 of the following paper

http://web.tecnico.ulisboa.pt/~mcasquilho/compute/com/,ellips/PerimeterOfEllipse.pdf
i.e.
\[P=\pi(a+b) \left[ 1+(\frac{1}{2})^2h^2+(\frac{1}{2.4})^2h^4+(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 8})^2h^8+ \dots\right],\]
where $h=\frac{a-b}{a+b}$.

On the other if one computes the taylor series expansion of the function
\[f(h)=\pi (a + b) \left(1 + \frac{3h^2}{10 + \sqrt{4 - 3h^2}}\right)\]
near $h=0$ we get
\[f(h)=\pi(a+b) \left[ 1+(\frac{1}{2})^2h^2+(\frac{1}{2.4})^2h^4+(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 8})^2h^8+ \dots\right].\]
Ineed $f(h)$, Ramanujan approximation for circumference of an ellipse, is an $O(h^{10})$ approximation of the circumference.

Erdos

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Erdos

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$Deriving Ramanujan's Ellipse approximation$

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