4

Take:

$$a_0=x,~~~~b_0=y$$

$$a_{n+1}=\frac{a_n+\sqrt{a_nb_n}}{2},~~~~b_{n+1}=\frac{b_n+\sqrt{a_nb_n}}{2}$$

Then we obtain as a limit the logarithmic mean of $x,y$:

$$\lim_{n \to \infty} a_n=\lim_{n \to \infty} b_n=\frac{x-y}{\ln x-\ln y}$$

I don't know how to prove this. But I do know that numerically it fits really well.

In fact, the best approximation is obtained if we take geometric mean of $a_n,b_n$:

$$x=5,~~~~y=3$$

$$\begin{array}( n & \sqrt{a_nb_n} & \frac{x-y}{\ln x-\ln y} \\ 4 & \color{blue}{3.915}0640985032 & 3.9152303779424 \\ 10 & \color{blue}{3.915230}33734566 & 3.9152303779424 \\ 20 & \color{blue}{3.9152303779424} & 3.9152303779424 \end{array}$$

The convergence rate can be approximated by:

$$\frac{a_{n+1}-b_{n+1}}{a_n-b_n}=\frac{1}{2}$$

This seems like a very simple way to compute logarithms, for example:

$$x=2,~~~~y=1$$

$$\ln2=\lim_{n \to \infty}\frac{1}{\sqrt{a_nb_n}}$$

How do I prove that the limit of this sequence is really the logarithmic mean?

Edit

It turns out this algorithm is mentioned in (at least) two papers by B. C. Carlson as early as 1971:

https://www.jstor.org/stable/2317088

https://www.jstor.org/stable/2317754

Still, if someone can provide their own proof, I would be grateful.

Martin Sleziak
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Yuriy S
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  • Similar sequence: https://en.wikipedia.org/wiki/Arithmetic%E2%80%93geometric_mean I bet one can solve it in a similar fashion, i.e. find some elliptic (or similar) integral that has the property that $I(x,y) = I(x + \sqrt{xy}/2, y + \sqrt{xy}/2)$. – Winther Jun 14 '16 at 14:35
  • @Winther, there are at least three integral forms for the logarithmic mean, so yes – Yuriy S Jun 14 '16 at 14:45
  • I'd simply prove that the logarithmic mean of $a_{n+1}$ and $b_{n+1}$ is the same as that of $a_n$ and $b_n$. – Ivan Neretin Jun 14 '16 at 14:53
  • @IvanNeretin, seems sensible, thank you for the idea – Yuriy S Jun 14 '16 at 15:53

1 Answers1

5

You have several exact identities, such as $$ a_{n+1}-b_{n+1}=\frac{a_n-b_n}2\implies a_n-b_n=2^{-n}(a_0-b_0), \\ \frac{a_{n+1}}{b_{n+1}}=\sqrt{\frac{a_n}{b_n}}\implies \frac{a_n}{b_n}=\left(\frac{a_0}{b_0}\right)^{2^{-n}} $$ The first tells us that if one of the sequences has a limit the other has the some limit. In combination $b_n$ can be eliminated to get, via mean value theorem, or $N(\sqrt[N]x-1)\to \ln x$ for $N\to \infty$, $$ a_n=\frac{2^{-n}(a_0-b_0)}{1-\left(\frac{b_0}{a_0}\right)^{2^{-n}}} =\frac{a_0-b_0}{\ln a_0-\ln b_0+O(2^{-n})} $$

Lutz Lehmann
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