TITLE: What comes after geometric mean?
QUESTION [3 upvotes]: The formula for computing the geometric mean seems to be the arithmetic mean formula except with all operations "shifted up" by one in the hyperoperator chain.
While arithmetic mean is:
$$\frac{a+b+c+\cdots}{n}$$
In the geometric mean, the addition has been replaced with multiplication and division replaced with a root:
$$(a b c\cdots)^{1/n}$$
One might thus wonder whether this could be extended further, replacing the multiplication with an exponentiation and the root with a "super-root," the inverse of a tetration.
Something like:
$$\left(a^{b^{c^{\cdots}}}\right)_n$$
Where the subscript $n$ stands for $n$th super-root.
Of course, this is problematic for the reason that exponentiation is not commutative; nonetheless, is there any such generalization of means and what significance could it hold?

REPLY [0 votes]: The idea of an exponent-based geometric mean has pretty much been in exile since people realized that exponents aren't commutative. But there still is a mathematically valid way to handle this.
The only way to counter the exponent's non-commutation in this case is to create two exponential means, one which goes in ascending order, and the other going in descending order. This virtually gets rid of the issue that exponents don't commute because they are now ordered in either ascending or descending order, so the two means are defined rigorously. 
Assuming this expression from Wikipedia is used to find the second super-root, the two means can be used successfully.  For example, the ascending exponential mean of $2$ and $3$ is $\sqrt{2^3}_s = \sqrt{8}_s ≈ 2.38842$ where $\sqrt{n}_s$ is the super-square root of $n$. The descending mean of $2$ and $3$ is equal to $\sqrt{3^2}_s = \sqrt{9}_s  ≈ 2.450953928$.
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