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derivation of geometric mean as the limit of the power mean
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(Derivation)
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Fix $x_1, x_2, \ldots, x_n \in \mathbb{R}^+$ . Then let$$ \mu(r) := \left(\frac{x_1^r+\cdots+x_n^r}{n}\right)^{1/r}.$$
For $r\neq 0$ , by definition $\mu(r)$ is the $r$ th power mean of the $x_i$ . It is also clear that $\mu(r)$ is a differentiable function for $r\neq 0$ . What is $\lim_{r\to 0} \mu(r)$ ?
We will first calculate $\lim_{r\to 0} \log\mu(r)$ using l'Hôpital's rule.
It follows immediately that$$ \lim_{r\to 0} \left(\frac{x_1^r+\cdots+x_n^r}{n}\right)^{1/r} = \sqrt[n]{x_1\cdots x_n}.$$
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"derivation of geometric mean as the limit of the power mean" is owned by Mathprof. [ full author list (3) | owner history (2) ]
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Cross-references: differentiable function, clear, power mean
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This is version 5 of derivation of geometric mean as the limit of the power mean, born on 2004-04-05, modified 2006-09-16.
Object id is 5741, canonical name is DerivationOfHarmonicMeanAsTheLimitOfThePowerMean.
Accessed 9048 times total.
Classification:
| AMS MSC: | 26D15 (Real functions :: Inequalities :: Inequalities for sums, series and integrals) |
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Pending Errata and Addenda
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![$\displaystyle = \log \sqrt[n]{x_1\cdots x_n}.$](http://images.planetmath.org:8080/cache/objects/5741/js/img6.png "$\displaystyle = \log \sqrt[n]{x_1\cdots x_n}.$"){kind=small}