# Modified AGM: $a_{n+1}=\frac{a_n+b_n}{2}, \quad b_{n+1}=a_n+b_n-\sqrt{a_n b_n}$

The idea is as follows: generate a sequence from two numbers by subtracting their means from their sum, for example arithmetic and geometric means:

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

$$b_{n+1}=a_n+b_n-\sqrt{a_n b_n}$$

$$\lim_{n \to \infty} a_n=\lim_{n \to \infty} b_n=L(a_0,b_0)$$

This one looks similar to arithmetic-geometric mean, but it's not the same. I tried computing it for several pairs of numbers, but didn't get any known constant from ISC or WA.

$$L(1,2)=1.54319140418204860465015096064$$

$$L(1,\sqrt{2})=1.21607308161790985508121278229$$

etc.

The convergence is great because:

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

Have you heard about this kind of iterated means in general and the titular one in particular? Is there any closed form known?

A little clarification. This is actually equivalent to adding the 'errors' to initial values, i.e. at each step we take the signed difference between one of the numbers and one of the means and add it to another number.

Edit

If we denote:

$$x_n=\frac{a_n}{b_n}$$

Then we obtain another definition of this recurrence:

$$x_{n+1}=\frac{1}{2} \frac{1+x_n}{1+x_n-\sqrt{x_n}}$$

$$a_{n+1}=a_0 \prod_{k=0}^n \frac{1}{2} \left(1+\frac{1}{x_k} \right)$$

This way the problem reduces to finding the explicit form for $x_n$. This is analogous to the Legendre form of the arithmetic-geometric mean.

I wouldn't hold out much hope for an explicit form. Getting rid of the square root, we obtain:

$$x_nx_{n+1}^2=(x_n+1)^2(x_{n+1}-1/2)^2$$

Also $x_n<1$ for all $n>0$. Which might point to a trigonometric substitution.

Or, alternatively:

$$y_n=\frac{b_n}{a_n}$$

$$y_{n+1}=2- \frac{2\sqrt{y_n}}{1+y_n}=2-\cfrac{2\sqrt{y_n}}{3-\cfrac{2\sqrt{y_{n-1}}}{3-\cfrac{2\sqrt{y_{n-2}}}{3-\cfrac{2\sqrt{y_{n-3}}}{1+y_{n-3}}}}}$$

$$a_{n+1}=a_0 \prod_{k=0}^n \frac{1}{2} \left(1+y_k \right)$$