Formula for the perimeter of an ellipse

Question

So despite being rather amateur when it comes to this level of math, I tried my hand at this and am quite satisfied with the answer. I've checked the numbers and it's very close to Ramanujan's approximation (never off by more than half of one percent), but I'd really love a second opinion or two.

For semi- axis $a$ and $b$:$$P\approx4b\sqrt[{\large n}]{\frac{a^n}{b^n}+1}; n={\frac{\log(2)}{\log({\frac{\pi}{2}})}}; b\neq0$$ Ramanujan's approximation: $$P_R\approx\pi(a+b)(1+\frac{3h}{10+\sqrt{4-3h}}; h=\frac{(a-b)^2}{(a+b)^2}$$

It is most off when the ratio of a/b is about 5: when a=5 and b=1, $P=21.0861$, and $P_R=21.0100$: a difference of .362%.

Answer 1

These two are even more accurate (the Pade approximation has a maximum error of .02%):

$$ \text{Ramanujan-Cantrell: Perimeter} = \pi(a+b)\left( 1 + \frac{3h}{10 + \sqrt{4 - 3h}} + \left( \frac{4}{\pi} - \frac{14}{11} \right) h^{12}\right) $$

Finally we have a rational function approximation known as a Padé approximation (named after French mathematician Henri Padé).

$$ \text{3rd Padé Approximation: Perimeter } =\pi(a+b)\left( \frac{135168 - 85760h - 5568h^2 + 3867h^3}{135168 - 119552h + 22208h^2 - 345h^3} \right) $$

Answer 2

Let $b=k\,a$ and you want to write $$E\left(1-k^2\right)\sim k \left(1+\left(\frac{1}{k}\right)^n\right)^{\frac{1}{n}}$$ Using Taylor series around $k=1$ $$E\left(1-k^2\right)=\frac{\pi }{2}+\frac{\pi }{4} (k-1)+O\left((k-1)^2\right)$$ $$k \left(1+\left(\frac{1}{k}\right)^n\right)^{\frac{1}{n}}=2^{\frac{1}{n}}+ 2^{\frac{1}{n}-1}(k-1)+O\left((k-1)^2\right)$$ Comparing the constant term gives $$2^{\frac{1}{n}}=\frac{\pi }{2} \quad \implies \quad n=\frac{\log (2)}{\log \left(\frac{\pi }{2}\right)} $$ which is your formula.

Using the next term of the series $$\text{lhs - rhs}=\frac{\pi \log \left(\frac{\pi ^3}{32}\right)}{32 \log\left(\frac{\pi }{2}\right)}(1-k)^2\sim -\frac{(1-k)^2}{146} $$

Making the model a bit more complex $$E\left(1-k^2\right)\sim k \left(1+\left(\frac{1}{k}\right)^n\right)^{\frac{1}{m}}$$ the same procedure gives $\large (!!)$ $m=n$; so, apparently, no way to improve this kind of formula (which is good).

The maximum absolute error is $0.0038$ and the maximum relative error is $0.362$% (as you wrote it).

Concerning the norm $$\Phi_1=\int_0^1 \Big(E(1-k^2)- \text{approximation}\Big)^2\,dk =4.956\times 10^{-6}$$

Edit

Using $$E\left(1-k^2\right)\sim k \left(a+\left(\frac{1}{k}\right)^n\right)^{\frac{1}{m}}$$ it seems that the model could be improved using $$a=\left(\frac{\pi }{2}\right)^m-1 \qquad \text{and} \qquad n=\frac {m\, \pi^m}{2^{m+1} }$$ $m$ being the implicit solution of $$2 m \left(\left(\frac{\pi }{2}\right)^m-1\right)=3\qquad \qquad m\sim \frac{17}{13-\sqrt[3]{6}}$$ the absolute error on $m$ being $4.65\times 10^{-8}$.

Using

$$\text{lhs - rhs}\sim \frac{(k-1)^3 }{1000}$$

Update

I do not remember where I saw the simple approximation $$E\left(1-k^2\right)\sim \frac{k \pi}{k+1} \csc \left(\frac{k \pi }{k+1}\right)$$ for which the maximum absolute error is $0.0034$ and the maximum relative error is $0.316$% (these are very similar to tour numbers).

Similarly $$\Phi_2=\int_0^1 \Big(E(1-k^2)- \text{approximation}\Big)^2\,dk =3.818\times 10^{-6}$$

But, what is amazing is that the errors are of opposite sign and the average $$E\left(1-k^2\right)\sim \frac 12\Bigg( k \left(1+\left(\frac{1}{k}\right)^n\right)^{\frac{1}{n}}+ \frac{k \pi}{k+1} \csc \left(\frac{k \pi }{k+1}\right)\Bigg)$$ seems to be interesting since the maximum absolute error becomes $0.00033$

$$\Phi_3=\int_0^1 \Big(E(1-k^2)- \text{approximation}\Big)^2\,dk =2.321\times 10^{-8}$$

$$\text{lhs - rhs}\sim -\frac{(k-1)^2 }{4372}$$