derivation of geometric mean as the limit of the power mean

Fix $x_1,x_2,\mathrm{\dots },x_n\in {ℝ}^{+}$. Then let

$$\mu (r):={\left(\frac{x_1^r+\mathrm{\cdots }+x_n^r}{n}\right)}^{1/r}.$$

For $r\ne 0$, 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 $r\ne 0$. What is ${lim}_{r\to 0}\mu (r)$?

We will first calculate ${lim}_{r\to 0}\log \mu (r)$ using l’Hôpital’s rule (http://planetmath.org/LHpitalsRule).

$\underset{r\to 0}{lim}\log \mu (r)$$=\underset{r\to 0}{lim}{\displaystyle \frac{\log \left(\frac{x_1^r+\mathrm{\cdots }+x_n^r}{n}\right)}{r}}$

$=\underset{r\to 0}{lim}{\displaystyle \frac{\left(\frac{x_1^r\log x_1+\mathrm{\cdots }+x_n^r\log x_n}{n}\right)}{\left(\frac{x_1^r+\mathrm{\cdots }+x_n^r}{n}\right)}}$

$=\underset{r\to 0}{lim}{\displaystyle \frac{x_1^r\log x_1+\mathrm{\cdots }+x_n^r\log x_n}{x_1^r+\mathrm{\cdots }+x_n^r}}$

$={\displaystyle \frac{\log x_1+\mathrm{\cdots }+\log x_n}{n}}$

$=\log \sqrt[n]{x_1\mathrm{\cdots }{x}_n}.$

It follows immediately that

$$\underset{r\to 0}{lim}{\left(\frac{x_1^r+\mathrm{\cdots }+x_n^r}{n}\right)}^{1/r}=\sqrt[n]{x_1\mathrm{\cdots }{x}_n}.$$

Title	derivation of geometric mean as the limit of the power mean
Canonical name	DerivationOfGeometricMeanAsTheLimitOfThePowerMean
Date of creation	2013-03-22 14:17:13
Last modified on	2013-03-22 14:17:13
Owner	Mathprof (13753)
Last modified by	Mathprof (13753)
Numerical id	8
Author	Mathprof (13753)
Entry type	Derivation
Classification	msc 26D15
Related topic	LHpitalsRule
Related topic	PowerMean
Related topic	WeightedPowerMean
Related topic	ArithmeticGeometricMeansInequality
Related topic	ArithmeticMean
Related topic	GeometricMean
Related topic	DerivationOfZerothWeightedPowerMean