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Let's introduce the following iterative procedure. Take two numbers $x_0$ and $y_0$.

$$a_0=\frac{x_0+y_0}{2}~~~~~~~~~~~b_0=\frac{2x_0y_0}{x_0+y_0}$$

$$x_1=\frac{x_0+a_0+b_0}{3}~~~~~~~~~~~y_1=\frac{y_0+a_0+b_0}{3}$$

$$\lim_{n \to \infty} x_n=\lim_{n \to \infty} y_n=P(x_0,y_0)$$


The proof of convergence. Let $x_0 \geq y_0$. Writing explicit expression for $x_1$ we see:

$$x_1= \frac{3x_0^2+y_0^2+8x_0y_0}{6(x_0+y_0)} \leq x_0$$

$$3x_0^2+y_0^2+8x_0y_0 \leq 6(x_0+y_0)x_0$$

$$y_0^2+2x_0y_0 \leq 3x_0^2$$

The above inequality is proven. In the same way we prove:

$$x_0 \geq x_1 \geq y_1 \geq y_0$$

The convergence rate:

$$x_1-y_1=\frac{1}{3}(x_0-y_0)$$

$$x_n-y_n=\frac{1}{3^n}(x_0-y_0)$$

A direct proof of the convergence since $\frac{1}{3} < 1$.


The generalization. Unlike AGM, this mean is easy to generalize for any $n$ numbers:

$$a_0=\frac{1}{n} \sum_{k=1}^{n} x_{k0}~~~~~~~~~~~b_0=\frac{n}{\sum_{k=1}^{n} 1/x_{k0}}$$

$$x_{k1}=\frac{x_{k0}+a_0+b_0}{3}$$


Everyone heard of Arithmetic-Geometric mean of two numbers, which is very useful in application to elliptic integrals. On the other hand, Arithmetic-Harmonic mean of two numbers if we get it the same way, is just equal to their Geometric mean, so it's not very interesting (only for computing square roots).

This mean $P(x,y,\dots)$, on the other hand is not equal to GM (actually, it seems to be always larger). For example:

$$P(1,2)=1.4372292891$$

$$P(1,2,3)=1.8638499699$$

The question is: Are there any publications about this mean? Does it have any applications (for computation of integrals, constants, etc)? If you haven't heard anything about it, what would you recommend me to do next to research it?

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For the case of two numbers, the recurrence can be made very simple. Using:

$$x_{n+1}= \frac{3x_n^2+y_n^2+8x_ny_n}{6(x_n+y_n)} $$

$$y_{n+1}= \frac{x_n^2+3y_n^2+8x_ny_n}{6(x_n+y_n)} $$

And:

$$x_n-y_n=\frac{1}{3^n}(x_0-y_0)=\frac{\delta}{3^n}$$

We set a new variable:

$$p_n=x_n+y_n$$

$$x_ny_n=\frac{1}{4} \left(p_n^2-(x_n-y_n)^2 \right)=\frac{1}{4} \left(p_n^2-\frac{\delta^2}{3^{2n}} \right)$$

And obtain a simple one-variable recursion:

$$p_{n+1}=p_n-\frac{\delta^2}{3^{2n+1} p_n}$$

$$\delta=x_0-y_0, \qquad p_0=x_0+y_0$$

$$\lim_{n \to \infty} x_n=\lim_{n \to \infty} y_n=\frac{1}{2} \lim_{n \to \infty} p_n$$

The convergence of this sequence is about two times faster than the original.

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