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What is this average called? I tried Googling it but couldn't find anything about it. It's something between an arithmetic mean and a geometric mean, by which I mean you iterate both averages for two numbers until they converge. To find that average of $a_0$ and $b_0$ you'd do this:

$$ a_{n+1} = \frac{a_{n} + b_{n}}{2}\\ b_{n+1} = \sqrt{a_{n} * b_{n}} $$

And repeat until the values converge. For instance with $a_{0}=1$ and $b_{0}=64$ it eventually converges towards around the value 18.1285.

Also is there a way to directly calculate that average in a non-iterative way?

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    $\begingroup$ This is arithmetic-geometric mean $M(a, b) $ $\endgroup$ – servabat Feb 12 '15 at 15:12
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    $\begingroup$ According to the wikipedia article, there is a (slightly) more direct way to calculate it as $$ \frac{\pi}{2}\Big/\int_0^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{a_0^2\cos^ 2\theta + b_0\sin^2 \theta}} $$ but since the sequences $a_n, b_n$ converge quite fast (roughly doubling the number of correct digits each time), I would use them for most practical purposes (assuming you are in a context where square roots are easily calculated, of course) $\endgroup$ – Arthur Feb 12 '15 at 15:20
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    $\begingroup$ See en.wikipedia.org/wiki/Arithmetic%E2%80%93geometric_mean $\endgroup$ – Crostul Feb 12 '15 at 15:24
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This is too long for a comment

Using the information given in comments and the content of the Wikipedia article, the arithmetic-geometric mean write $$M(a,b)=\frac{\pi b_0}{2 K\left(1-\frac{a_0^2}{b_0^2}\right)}$$ where appears the elliptic integral of the first kind .

Considering the case where $a_0 <<b_0$, we can derive an asymptotic formula $$M(a,b) \approx\frac{\pi b_0}{\log \left(\frac{16 b_0^2}{a_0^2}\right)}\Big[1-4\frac{ \log \left(\frac{16 b_0^2}{a_0^2}\right)-2}{\left(\frac{16 b_0^2}{a_0^2}\right) \log \left(\frac{16 b_0^2}{a_0^2}\right)}+\cdots\Big]$$ which appears to be a very good approximation. For $a=1,b=64$ as in the post, the exact value is $18.12853351$ while the given approximation leads to $18.12853358$.

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