Iterated means $a_{n+1}=\sqrt{a_n \frac{b_n+c_n}{2}}$, $b_{n+1}$ and $c_{n+1}$ similar, closed form for general initial conditions?

Question

For every nonnegative $(a_0,b_0,c_0)$, consider

$$a_{n+1}=\sqrt{a_n \frac{b_n+c_n}{2}},\quad b_{n+1}=\sqrt{b_n \frac{c_n+a_n}{2}},\quad c_{n+1}=\sqrt{c_n \frac{a_n+b_n}{2}}$$

$$M(a_0,b_0,c_0)=\lim_{n \to \infty} a_n=\lim_{n \to \infty} b_n=\lim_{n \to \infty} c_n$$

This simple three term iterated mean gives several interesting closed forms for particular cases (confirmed to $14$ digits so far):

$$M(1,1,2)=\frac{3^{3/4}}{\sqrt{\pi}}=1.28607413715749$$

$$M(1,1,\sqrt{2})=\frac{2}{\sqrt{\pi}}=1.12837916709551$$

$$M(1,1,\sqrt{3})=\frac{2^{3/4}}{\sqrt{\arccos(-1/3)}}=1.21670090936316$$

$$M(1,1,\sqrt{3}/2)=\frac{1}{\sqrt{\ln 3}}=0.95406458200000$$

These particular closed forms (if true) imply that the general closed form for $M(a_0,b_0,c_0)$ should also exist and involves the known elementary (or at least special) functions.

I have no idea how to find it so far and would appreciate the help.

It seems that for $M(1,1,x)$ the closed form should involve $\frac{1}{\sqrt{\arccos(f(x))}}$. Edit. Except in one case it has logarithm.

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Since this mean can be considered yet another generalization of the Arithmetic Geometric Mean for three numbers, I link related questions: 1, 2.

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To anyone interested, there is a companion sequence, which corresponds to the square roots of the $a_n,b_n,c_n$ of the titular sequence:

$$a_{n+1}=\frac{\sqrt{a_nb_n}+\sqrt{a_nc_n}}{2},\quad b_{n+1}=\frac{\sqrt{b_na_n}+\sqrt{b_nc_n}}{2},\quad c_{n+1}=\frac{\sqrt{c_na_n}+\sqrt{c_nb_n}}{2}$$

The limit of this sequence $M'(a_0,b_0,c_0)$ corresponds to the previous one:

$$M'(a_0,b_0,c_0)=(M(\sqrt{a_0},\sqrt{b_0},\sqrt{c_0}))^2$$

For example:

$$M'(1,1,2)=(M(1,1,\sqrt{2}))^2=\frac{4}{\pi}$$

Answer 1

Analogous to the Schwab-Borchardt mean:

\begin{align*} B(a,b) &= \frac{\sqrt{b^{2}-a^{2}}}{\cos^{-1} \frac{a}{b}} \\ &= B\left( \frac{a+b}{2}, \sqrt{\frac{(a+b)b}{2}} \right) \end{align*}

which can also be obtained by the iteration: $$ \left \{ \begin{array}{rcl} a_{n+1} &=& \frac{a_{n}+b_{n}}{2} \\ b_{n+1} &=& \sqrt{a_{n+1} b_{n}} \end{array} \right.$$

Now your mean is

\begin{align*} M(x,x,y) &= D(x,y) \\ &= D\left( \sqrt{\frac{x(x+y)}{2}} , \sqrt{xy} \right) \\ &= \frac{\sqrt[4]{x^{2}(y^{2}-x^{2})}}{\sqrt{\cos^{-1} \frac{x}{y}}} \end{align*}

Answer 2

let $L(x) = M(1,1,x)^2$. Using properties of $M$, we have that $L$ is the unique function on $\Bbb R_{>0}$ continuous at $1$ satisfying $L(1) = 1$ and the "recurrence relation" :

$L(x) = M(\frac{\sqrt{1+x}}2,\frac{\sqrt{1+x}}2,\sqrt x)^2 = \frac{1+x}4M(1,1,\sqrt\frac{2x}{1+x})^2 = \frac{1+x}4 L(\sqrt\frac{2x}{1+x})$

$L$ is continuous at $1$ because $M(1,1,x)$ is between $1$ and $x$.
Such a function is unique because if there was another one $f$, then $f$ doesn't vanish, and $L/f$ is continuous at $1$ and satisfies $(L/f) (x) = (L/f) (\sqrt\frac{2x}{1+x})$. Since iterating $x \mapsto \sqrt\frac{2x}{1+x}$ on any positive real gives a sequence that converges to $1$, we must have $(L/f) (x) = \lim_{y \to 1} (L/f) (y) = 1/1 = 1$.

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Let $f$ be the function defined by $f(x) = \frac{\sqrt{x^2-1}}{\arccos \frac 1x}$ if $x > 1$ and $f(1) = 1$ (it should also actually make sense for $0<x<1$ if you take the right pair of complex square roots and arccos)

Some trigonometric identities and some algebra shows that $f$ satisfies the recurrence relation of $L$ .

so we only need to show that $f$ is continuous at $1$ to prove that $L=f$.

Since the map $h \mapsto \cos(\sqrt h)$ is differentiable on the right at $0$ (with derivative $-1/2$), its inverse map $y \mapsto (\arccos y)^2$ is differentiable on the left at $1$ with derivative $-2$.

Now, for $x>1$, $f(x)^2 = \frac {(x-1)(x+1)}{(\arccos \frac 1x)^2} = - \frac {\frac 1x-1}{(\arccos \frac 1x)^2}x(x+1)$ and by definition of the derivative, the big fraction converges to $-1/2$ when $x \to 1$, so the whole thing converges to $-(-1/2)\cdot 1 \cdot 2 = 1$. Therefore $f^2$ is continuous at $1$, and then so is $f$.

This proves that $L(x) = f(x)$ for $x \ge 1$.