Taking the means of numbers repeatedly

Question

Let $A_1 = \{x_1,x_2,x_3,x_4\}$ be a set of four positive real numbers. The sets $A_i, i\geq 2$ are made up of four real numbers defined from the arithmetic mean, geometric mean, harmonic mean and root mean square, as shown below.

Let $A_2 = \{ \text{AM} (A_1),\text{GM}(A_1),\text{HM}(A_1),\text{RMS}(A_1) \}$

Let $A_3 = \{ \text{AM} (A_2),\text{GM}(A_2),\text{HM}(A_2),\text{RMS}(A_2) \}$

$ \vdots$

Let $A_{i+1} = \{ \text{AM} (A_i),\text{GM}(A_i),\text{HM}(A_i),\text{RMS}(A_i) \}$

I was wondering if there was anything special as this process is repeated. Do the elements of the set converge towards a specific number? I was thinking it would converge to SOME number between the AM and GM since $RMS \geq AM \geq GM \geq HM$, but this is as far as I could go.

Answer 1

I can't give a complete answer, but here's what I did:
(note that in your question, you put $AM \geqslant GM \geqslant HM \geqslant RMS$, but the RMS is actually supposed to be the greatest, as in $RMS \geqslant AM \geqslant GM \geqslant HM$)
To experiment, I took 3 sets of numbers $A_1 = \{2, 6, 11, 27\}, B_1 = \{1, 2, 3, 4\}$, and $C_1 = \{182.26, 5\sqrt{2}, 192403, 12\sin(1)\}$, put them in excel, and crunched the numbers.
As it turns out, the numbers do converge to a specific number unique to each set. Some converge faster than others, but they do converge and at that relatively quickly. Here's a picture for clarity:

So, the three sets do converge to a single number, highlighted in red when all the values reached an agreement to 9 decimal places. As you can see, the number is always between the AM and GM, not the GM and HM. (Although I suspect your reasoning was correct in thinking that it would converge to a number in the 'middle' of the inequalities.)
I don't have any concrete ideas on how to give a proof of this fact, but it might come from this reasoning:

1. Given a set of 4 positive numbers in a set $A_1$, we compute their AM, GM, HM, and RMS, and let them be the four values in the set $A_2$.
2. Since the four numbers in $A_2$ satisfy the RMS-AM-GM-HM inequality, and every average must be in between its greatest and least values (the largest and smallest elements in $A_1$), then $\max{A_1} \geqslant 1stRMS \geqslant 1stAM \geqslant 1stGM \geqslant 1stHM \geqslant \min{A_1}$.
3. In $A_3$, by the same reasoning as in step 2, we have that the maximum of $A_2$, which is the 1stRMS, will be greater than the 2ndRMS, which is in turn greater than the 2ndAM... which is larger than the minimum of $A_2$, which is the 1stHM.
4. We can keep on 'nesting' the inequalities in each other with each repetition, and in doing this we are squeezing the means in between increasingly close numbers, which will make it converge.

Cheers!

Answer 2

Late for the party but here goes: Yes,there is convergence. For just arithmetic and geometric mean this is classical (check the AGM method). For more means there is an article by Krause

Krause, Ulrich, Compromise, consensus, and the iteration of means. Elem. Math. 64 (2009), no. 1, 1–8.

which proves convergence in a wider setting. You can even change the used means from iteration to iteration and still obtain convergence as my brother and me proved here

Lorenz, Jan; Lorenz, Dirk A., On conditions for convergence to consensus. IEEE Trans. Automat. Control 55 (2010), no. 7, 1651–1656.