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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?

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  • $\begingroup$ Other problems besides non commutativity are that the $n$-th super root does not always exists (even for positive arguments) and that it is in general multivalued (and even two different positive super roots can occur). $\endgroup$ – Dirk Dec 9 '15 at 11:37
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    $\begingroup$ An idea: $s_0 = \frac{1}{n} \sum_{i=1}^n a_i$, $s_1 = \exp \left(\frac{1}{n} \sum_{i=1}^n \ln a_i \right) = \sqrt[n]{\prod_{i=1}^n a_i}$, $s_2 = \exp \exp \left(\frac{1}{n} \sum_{i=1}^n \ln \ln a_i \right)$, $s_k = \exp^k \left(\frac{1}{n} \sum_{i=1}^n \ln^k a_i \right)$ ($\exp^k$ and $\ln^k$ denote iterated functions). There are always some complex values, but it may be hard to find the main value of $\ln^k$ below $\exp^{k-1} 0$. $\endgroup$ – BartekChom Dec 9 '15 at 12:01
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There is a different sort of mean: $$\left(\frac{a^p+b^p+c^p+\cdots}n\right)^{1/p}$$ It includes:

  • the usual mean, when $p=1$
  • the 'root-mean-square', when $p=2$
  • the geometric mean, when $p=0$ (it turns into $1^{\infty}$, which has no particular value, so you have to do calculus, and it becomes the geometric mean)
  • the harmonic mean, when $p=-1$
  • the maximum, when $p=\infty$
  • the minimum, when $p=-\infty$
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  • $\begingroup$ Interesting, never seen that before! Does this generalized expression have a name? $$$$ If you were to use this to extrapolate something after the geometric mean, what value would you use for $p$? $\endgroup$ – 1110101001 Nov 28 '15 at 22:03
  • $\begingroup$ There is a page on it in Wikipedia, en.wikipedia.org/wiki/Generalized_mean $\endgroup$ – Empy2 Nov 28 '15 at 22:07

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