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It seems like taking the n-th root of the product of n values is analogous to the traditional form of averaging. For example:

$$ \frac{5+5+5}{3} = 5 = \sqrt[\leftroot{0}\uproot{0}3]{(5*5*5)} $$

But they do not always produce the same results

$$ \frac{1+2+3}{3} = 2 $$ $$ \sqrt[\leftroot{0}\uproot{0}3]{(1*2*3)} = 1.817... $$

Also, the n-th root "average" cannot handle negative numbers. Anyway, what are the benefits of this latter form of averaging versus the standard way?

(Context: I'm in a course where "perplexity" was intuitively defined as the "average amount of surprise". It made me wonder why we don't just use the standard way of averaging.)

jds
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    Relevant: https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means (You are wondering about the relation between arithmetic and geometric means). – Clement C. Oct 27 '16 at 18:04
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    @clementc. I believe that the OP is asking for an application for which a geometric mean is appropriate and more meaningful in some sense than an arithmetic one. – Mark Viola Oct 27 '16 at 18:20
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    The geometric mean is less sensitive for exceptional values. An extreme example : $$\frac{10+10+10+10+10^9}{5}=2\cdot 10^8+8$$ and $$(10\cdot 10\cdot 10\cdot 10\cdot 10^9)^{1/5}\approx398.1$$ – Peter Oct 27 '16 at 18:30

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The geometric mean can be thought of as $e^y$ where $y$ is the average of the natural logarithms of your values. That is $$\left(\prod_{i=1}^n x_i\right)^\frac 1 n = e^{ {\left( \sum_{i=1}^n \ln x_i \right)} \over n}$$Thus this sort of average is useful when your values can be thought of as powers of some base, and you want to find the average value of the exponent.

Gabriel Burns
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