nth-order generalizations of the arithmetic-geometric mean

Question

The arithmetic-geometric mean,

$a_{k+1}=\frac{a_k+b_k}{2} \quad b_{k+1}=\sqrt{a_k b_k}$

is one of the celebrated discoveries of Gauss, who found out that it is equivalent to computing a (complete) elliptic integral (which is a special case of the Gauss hypergeometric function ${}_2 F_1$).

I have been wondering if nth-order generalizations of the iteration,

$a_{k+1}=\frac{a_k+b_k}{n} \quad b_{k+1}=\sqrt[n]{a_k b_k}$

have ever been systematically studied. I've seen this paper (Wayback Machine) by Borwein, but have had trouble searching for other papers. In particular, I'm interested if the coupled sequences also have a common limit, and if so, whether the limit is expressible as a hypergeometric function (or generalizations like those of Appell or Lauricella).

Another possible generalization I thought involves $n$ variables and makes use of the elementary symmetric polynomials. To use $n=4$ as an example:

$a_{k+1}=\frac{a_k+b_k+c_k+d_k}{4}$

$b_{k+1}=\sqrt{\frac{a_k b_k+a_k c_k+a_k d_k+b_k c_k+b_k d_k+c_k d_k}{3}}$

$c_{k+1}=\sqrt[3]{\frac{{a_k b_k c_k}+{a_k b_k d_k}+{a_k c_k d_k}+{b_k c_k d_k}}{2}}$

$d_{k+1}=\sqrt[4]{a_k b_k c_k d_k}$

Would these four sequences (and in general the $n$ sequences) tend to a common limit $F(a_0,b_0,c_0,d_0,\dots)$ like in the $n=2$ case, and if so, are they expressible in terms of known functions?

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EDIT

Taking into account Darsh Ranjan's comments, I realized that what I should be looking at instead is the generalization whose denominators are binomial coefficients (thus, the general form $\sqrt[j]{\frac{e_j}{\binom{n}{j}}}$, for $j=1\dots n$ where $e_j$ is the jth elementary symmetric polynomial). The case $n=4$ now looks like

$a_{k+1}=\frac{a_k+b_k+c_k+d_k}{4}$

$b_{k+1}=\sqrt{\frac{a_k b_k+a_k c_k+a_k d_k+b_k c_k+b_k d_k+c_k d_k}{6}}$

$c_{k+1}=\sqrt[3]{\frac{{a_k b_k c_k}+{a_k b_k d_k}+{a_k c_k d_k}+{b_k c_k d_k}}{4}}$

$d_{k+1}=\sqrt[4]{a_k b_k c_k d_k}$

So, still the same question: is there a common limit, and if so, is the limit expressible in terms of known functions?

Answer 1

You want Maclaurin's inequality. Given $n$ positive numbers $a_1, a_2,\dots,a_n$, write $$ (x+a_1)(x+a_2)\cdots(x+a_n) = x^n + S_1x^{n-1} + \cdots + S_{n-1}x + S_n, $$ so $S_i$ is the $i$-th elementary symmetric function of $a_1,\dots,a_n$. For $i = 1,\dots,n$, set $A_i = S_i/\binom{n}{i}$. When $n = 2$, $A_1 = (a_1+a_2)/2$ and $A_2 = a_1a_2$. Maclaurin's inequality is that $$ A_1 \geq \sqrt{A_2} \geq \sqrt[3]{A_3} \geq \cdots \geq \sqrt[n]{A_n}, $$ where the inequality signs are all strict unless $a_1,\dots,a_n$ are all equal. The inequality of the outer terms, $A_1 \geq \sqrt[n]{A_n}$, is the arithmetic-geometric mean inequality for $n$ positive numbers.

From a list of $n$ positive numbers $a_1,\dots,a_n$ we have produced another list of $n$ positive numbers $A_1,\sqrt{A_2},\dots,\sqrt[n]{A_n}$. The construction can be repeated.

Theorem: All the terms in the list tend to the same limit.

Off the top of my head I can't recall a reference where this is proved. It was studied by Meissel in 1875 for $n = 3$.

For example, if we start with the three numbers 1, 2, 3 then after 4 iterations the three numbers we get all look like 1.9099262335 to 10 digits after the decimal point.

[Edit: Here is a proof of the common limit, based on Will Sawin's first comment below to my answer. Order the numbers $a_1,\dots,a_n$ so that $a_1 \geq \cdots \geq a_n > 0$. By Maclaurin's inequality (or really just the arithmetic-geometric mean inequality) $A_1 \geq \sqrt[n]{A_n}$ and we will bound $A_1 - \sqrt[n]{A_n}$ from above in terms of $a_1 - a_n$ by bounding $A_1$ from above and $\sqrt[n]{A_n}$ from below using just $a_1$ and $a_n$. To bound $A_1$ from above,
$$ A_1 = \frac{a_1 + \cdots + a_n}{n} \leq \frac{(n-1)a_1 + a_n}{n} = a_1 - \frac{a_1 - a_n}{n} $$ and to bound $A_n$ from below we write $A_n = a_1\cdots a_n \geq a_n^n$, so $$ \sqrt[n]{A_n} \geq a_n. $$ Therefore $$ 0 \leq A_1 - \sqrt[n]{A_n} \leq \left(a_1 - \frac{a_1 - a_n}{n}\right) - a_n = \left(1 - \frac{1}{n}\right)(a_1 - a_n). $$ Start from an $n$-tuple $(a_1,a_2,\dots,a_n)$ which is ordered so that $a_1 \geq \cdots \geq a_n > 0$ and construct the $n$-tuple $(A_1,\sqrt{A_2},\dots,\sqrt[n]{A_n})$ and keep repeating this, which produces a sequence of $n$-tuples $(a_1^{(k)},a_2^{(k)},\dots,a_n^{(k)})$ for $k = 0,1,2,\dots$, where $a_i^{(0)} = a_i$. Let's look at the sequence of first coordinates $a_1^{(k)}$. An arithmetic mean of positive numbers is bounded above by the largest number, so $a_1 = a_1^{(0)} \geq a_1^{(1)} \geq a_1^{(2)} \geq \cdots > 0$. Therefore the sequence $a_1^{(k)}$ converges as $k \rightarrow \infty$. (The limit is positive because the sequence of last coordinates $a_n^{(k)}$ is non-decreasing and $a_1^{(k)} \geq a_n^{(k)} \geq a_n^{(0)} = a_n$ for all $k$.) The above calculation shows $$ 0 \leq a_1^{(k)} - a_n^{(k)} \leq \left(1 - \frac{1}{n}\right)(a_1^{(k-1)} - a_n^{(k-1)}), $$ so $0 \leq a_1^{(k)} - a_n^{(k)} \leq (1 - 1/n)^k(a_1 - a_n)$. Letting $k \rightarrow \infty$ we see the sequence of last coordinates $a_n^{(0)},a_n^{(1)},a_n^{(2)},\dots$ converges to the limit of the sequence of first coordinates $a_1^{(0)}, a_1^{(1)}, a_1^{(2)},\dots$. Since $a_1^{(k)} \geq a_i^{(k)} \geq a_n^{(k)}$, each intermediate sequence $a_i^{(0)},a_i^{(1)},a_i^{(2)},\dots$ converges to the same limit. ]

Answer 2

Gauss's hypergeometric formula for the AGM can also be interpreted in terms of a complete elliptic integral $\int_0^{\pi/2} \phantom. d\theta / \sqrt{a^2 \cos^2 \theta + b^2 \sin^2 \theta}$. There's a remarkable generalization to complete hyperelliptic integrals in genus 2 arising from a four-variable AGM: $$ (a,b,c,d) \mapsto \left( \frac14(a+b+c+d), \frac12\bigl(\sqrt{ab}+\sqrt{cd}\phantom.\bigr), \frac12\bigl(\sqrt{ac}+\sqrt{bd}\phantom.\bigr), \frac12\bigl(\sqrt{ad}+\sqrt{bc}\phantom.\bigr) \right) $$ (whose specialization $a=b$, $c=d$ recovers the usual AGM). One source available online is

Jarvis, Frazer: Higher genus arithmetic-geometric means, Ramanujan J. 17 (2008), 1–17.

Answer 3

You may want to check a book by J.M. Borwein and P.B. Borwein titled "Pi and the AGM". The convergence is quadratic in the sense that the number of settled digits doubles in every turn in the long run. AGM of 2 numbers leads to elliptic functions but for three or more numbers the problem seems open.